Find out all the rational roots of a rich term online. Equation in all mathematics. Rational root of rich terms. Horner's scheme. Chi є tse rational number

A rich term in the form of a variable x is called in a different way: anxn + an-1 xn-1 +. . . +a 1 x+a 0 de n is a natural number; an, an-1, . . . , a 1, a 0 - whether they are numbers, called coefficients of this polynomial. Virazi anxn, an-1 xn-1, . . . , a 1 x, a 0 are called members of the polynomial, and 0 is called an arbitrary member. an - coefficient at xn, an-1 - coefficient at xn-1 and so on. for example, the rich term 0x2 + 0x + 0 is null. From the record of the polynomial, it is clear that vin is added up from the number of members. Sounds like the term "rich member" (rich members). Sometimes a rich term is called a polynomial. This term resembles the Greek words πολι - rich and νομχ - member.

A rich member in the form of one change x is signified: . f (x), g (x), h (x) and so on, for example, as the first pointing more richly terms in terms of f (x), then you can write: f (x) = x 4+2 x 3+ (- 3) x 2 + 3/7 x + √ 2. 1. The rich term h (x) is called the largest sleeper of the rich terms f (x) and g (x), so it is possible to add f (x), g (x) and leather dilnik. 2. Rich term f(x) with coefficients from the field P of step n is called reducible over the field P, thus establishing rich terms h(x), g(x) Î P[x] of step less n such that f(x) = h( x)g(x).

This is the rich term f(x) = anxn+an-1 xn-1+. . . + a 1 x + a 0 і an≠ 0, then the number n is called the stage of the rich term f (x) (or it seems: f (x) is the n-th stage) and write Art. f(x) = n. And here an is called the senior coefficient, and anxn is the senior member of this polynomial. For example, if f (x) = 5 x 4 -2 x +3, then Art. f(x) = 4, senior coefficient - 5, senior term - 5 x4. The polynomial step is the largest of the numbers of its coefficients, the leading types of zero. The rich terms of the zero step are the whole numbers, which are the same as zero. the zero rich term of the step cannot be; rich term f(x) = a, where a is a number, not equal to zero, the maximum step is 0; step well be some other polynomial, which is more expensive to the greatest indicator of the step of change x, the coefficient at the next one is zero.

Rivnist of rich-membered. Two rich terms f(x) and g(x) are considered equal, even though their coefficients are equal at the same steps of the change x and free terms (equal їх відпровідні coefficients). f(x) = g(x). For example, rich terms f (x) \u003d x 3 + 2 x 2 -3 x + 1 і g (x) \u003d 2 x 23 x + 1 are not equal, the first of them has a coefficient at x3 more equal to 1, and the other has zero ( With the accepted intelligences, we can write: g (x) \u003d 0 x 3+2 x 2 -3 x + 1. In which case: f (x) ≠ g (x). x 2 -3 x + 5, s ( x) =2 x 2+3 x+5

And the axis of the rich term f 1 (x) \u003d 2 x 5 + 3 x 3 + bx + 3 і g 1 (x) \u003d 2 x 5 + ax 3 -2 x + 3 equally even if a = 3, but b = -2. Give the rich term f(x) = anxn+an-1 xn-1+. . . +a 1 x+a 0 is a number c. Number f(c) = ancn+an-1 cn-1+. . . +a 1 c+a 0 is called the value of the polynomial f(x) at x = c. In such a way, in order to know f (c), it is necessary to substantiate x and carry out the necessary calculations. For example, if f(x) = 2x3+3x2-x+5, then f(-2)=2(-2)3+(-2)2-(-2)+5=3. A rich member with different values \u200b\u200bof change x can be taken different values. The number is called the root of the polynomial f (x), so f (c) =0.

It is important to pay attention to the difference between two statements: "the rich term f(x) is equal to zero (otherwise, the rich term f(x) is zero)" and "the value of the polynomial f(x) at x=z is equal to zero". For example, the polynomial f (x) \u003d x 2 -1 is not equal to zero, vіn may be non-zero coefficients, like the value at x \u003d 1 is equal to zero. f(x) ≠ 0, and f(1) =0. Between the understandings of the equivalence of rich terms and the meaning of the rich term is the same close interrelationship. If two equal polynomials f(x) and g(x) are given, then їх are equal coefficients of equals, and, therefore, f(c) = g(c) for the skin number с.

Operations on polynomials Rich terms can be added, seen and multiplied according to the usual rules for expanding the arc and reduction of similar terms. With this, as a result, I again enter a rich member. Designated operations may have power: f (x) + g (x) = g (x) + f (x), f (x) + (g (x) + h (x)) = (f (x) + g (x)) + h(x), f(x) g(x) = g(x) f(x), f(x)(g(x) h(x)) = (f(x) g( x)) h(x), f(x)(g(x) + h(x)) = f(x) g(x) + f(x) h(x).

Let me give you two rich terms f(x) = anxn+an-1 xn-1+. . . +a 1 x+a 0, an≠ 0, i g(x)=bmxm+bm-1 xm-1+. . . +b 1 x+bm≠ 0. It was clear that Art. f(x)=n, and art. g(x) = m. If you multiply qi two polynomials, you will end up with a rich term of the form f(x) g(x)=anbmxm+n+. . . +a 0 b 0. Oskilki an≠ 0 and bn≠ 0, then anbm≠ 0, also, art. (f(x)g(x))=m+n. Sounds are loud and important.

Steps to add two non-zero rich terms to the sum of the steps of the multipliers, art. (f(x)g(x)) = st. f(x) +st. g(x). The senior member (coefficient) of the creation of two non-zero rich terms in order to add the senior members (coefficients) of the multipliers. A free member of the creation of two richly-membered members is worthy of the creation of free members of the joint multipliers. Steps of richly articulated f(x), g(x) and f(x) ±g(x) are associated with upcoming spivvіdnoshennia: art. (f (x) ± g (x)) ≤ max (st. f (x), st. g (x)).

The superposition of multiple terms f(x) and g(x) is called. rich term, which is denoted by f (g (x)), which can also go into the polynomial f (x) instead of x, substitute the polynomial g (x). For example, if f(x)=x 2+2 x-1 і g(x) =2 x+3, then f(g(x))=f(2 x+3)=(2 x+3)2 +2(2 x+3)-1=4 x 2+16 x+14, g(f(x))=g(x 2+2 x-1)=2(x 2+2 x-1)+ 3=2x2+4x+1. It can be seen that f(g(x)) ≠g(f(x)), that is a superposition of multiple terms f(x), g(x) and a superposition of multiple terms g(x), f(x) different. In this way, the operation of superposition does not have the power of displacement.

, Algorithm for underestimation and overflow For whether f(x), g(x) it is clear q(x) (privately) and r(x) (surplus), so that f(x)=g(x)q(x )+ r(x), and the steps r(x)

Dictionaries of a polynomial Dictionary of a rich term f(x) is a rich term g(x) such that f(x)=g(x)q(x). The largest bed of two richly-segmented The largest bed of richly-segmented f(x) and g(x) is such a double bed of d(x), which can be divided into any other bed of theirs.

Euclidean's algorithm (algorithm of the last sub-line) of the largest common diary of rich terms f(x) and g(x) Todi is the largest dilnik of f(x) and g(x).

Change others Solution: We know the GCD of these rich terms, fixing the Euclidean algorithm 1) x3 + 6 x2 + 11 x + 6 x3 + 7 x2 + 14 x + 8 8 x3 + 3 x2 + 2 x - x2 - 3 x - 2 - x - 4 4 x2 + 12 x + 8 0 Otzhe, rich term (- x2 - 3 x - 2) The result is under the banner of the polynomial of vіdomy.

Let's know the result of the subdivision of the number. x 3 + 6 x2 + 11 x + 6 - x2 - 3 x - 2 x3 + 3 x2 + 2 x - x - 3 3 x2 + 9 x + 6 0

Horner's scheme of dividing from an overly rich term f(x) into a non-zero rich term g(x) - ne means to reveal f(x) in the view f(x)=g(x) s(x)+r(x), de s(x) ) ) i r(x) -rich terms i or r(x) = 0, or st. r(x)

Rich segments, which stand at the left and right parts of its spіvvіdnoshennia, equal, and also, equal їhnі vіdpovіdni koefіtsіentsi. It is equal to them, having opened the bows in front and instilled similar limbs at the right part of the line of equanimity. Minus: a = bn-1, a-1 = bn-2 - cbn-1, a-2 = bn-3 - cbn-2, a 2 = b 1 - cb 2, a 1 = b 0 - cb 1, a 0 = r - cb 0 Virazimo їх іz otrimanih equalities: bn-1 = an, b n-2 = cbn-1 + an-1, b n-3 = cbn-2 + a n-2, b 1 = cb 2 + a 2, b 0 \u003d cb 1 + a 1, r \u003d cb 0 + a 0. We knew the formulas that can be used to calculate the coefficients of an odd private s (x) and excess r. With this, the charges are drawn up at the front of the table; it is called Horner's scheme.

Table 1. Coefficients f(x) c an bn-1 an-1 bn-2=cbn-1+ an-1 an-2 bn-3 = cbn-2+an-2 … … a 0 r = cb 0 + a 0 The coefficients s(x) are too much. In another row, near the first cell, write down the number c. Reshta clitin of the row is filled in, counting, one by one, the coefficients of the non-linear private s (x) and the excess r. At another client, write down the coefficient bn-1, which, as we have installed, is more expensive an.

The coefficient to stand at the skin offensive wall is calculated according to the following rule: the number c is multiplied by the number to stand at the front wall, and the number is added to the result, to stand above the wall, to be remembered. In order to remember, let's say, five clitin, to know to stand at her coefficient, it is necessary to multiply c by the number that is in the fourth clitin, and add to the result the number that stands above the fifth clitin. Let's divide, for example, the rich term f (x) \u003d 3 x 4 -5 x 2 + 3 x-1 into x-2 іz too much, Horner's scheme. When filling in the first row, the numbers of the scheme cannot be forgotten about the zero coefficients of the polynomial. So, the coefficients f(x) are the values of the numbers 3, 0, - 5, 3, - 1. Another thing to keep in mind is that the step of an incomplete private one is one lesser than the step of the rich term f(x).

Also, it seems to have been subdivided according to Horner's scheme: Table 2. 2 3 3 0 6 -5 7 3 17 -1 33 It is important to note that private s(x) =3 x 3+6 x 2+7 x+17 and surplus r=33. Respectfully, we have calculated the value of the polynomial f (2) =33. Now let's divide the very rich term f(x) into x + 2 іz too much. I have a vipadku with = -2. optional: Table 3. -2 3 3 0 -6 -5 7 3 -11 -1 21 As a result, f(x) = (x+2) (3 x 3 -6 x 2+7 x-11) +21 .

Root of polynomials Nehai с1, с2, …, сm - Different root of the polynomial f(x). Then f(x) is divisible by x-c1, then f(x) = (x-c1) s1(x). Let's pay for this equanimity x=c2. We subtract f(c2) = (c2-c1) s1(c2) i, so f(c2) =0, then (c2-c1) s1(c2) =0. Ale c2≠c1, then c2 -c1≠ 0, which means that s 1 (c 2) = 0. Also, c2 is the root of the polynomial s 1 (x). It shows that s1(x) is divisible by x-c2, so s1(x) = (x-c2) s2(x). Imagine subtracting virase for s 1 (x) y equal f (x) = (x-c 1) s 1 (x). May f(x) = (x-c1) (x-c2) s2(x). Having put in the rest of the equality x \u003d c3, to fix that f (c 3) \u003d 0, c3 c1, c3 c2, we assume that c3 is the root of the polynomial s 2 (x). So, s 2 (x) \u003d (x-c 3) s 3 (x), and then f (x) \u003d (x-c 1) (x-c 2) (x-c 3) s 3 (x) and so on. for roots that have been lost, c4, c5, ..., cm, mi, nareshti, f (x) = (x-c 1) (x-c 2) ... (x-cm) sm (x) is taken away, this is brought to a lower formula.

Since c1, c2, ..., cm is the different root of the polynomial f (x), then f (x) can be given by looking at f (x) \u003d (x-c 1) (x-c 2) ... (x-cm) sm (x). Sounds like an important consequence. Since c1, c2, ..., cm is the root of the polynomial f (x), then f (x) is divided by the polynomial (x-c1) (x-c2) ... (x-cm). The number of different roots of the non-zero polynomial f(x) is not greater than the lower step. True, since f(x) has no root, it is clear that the theorem is correct, more Art. f (x) ≥ 0. Let f (x) now have m roots c1, c2, ..., cm, moreover, all stinks are different. Just as f (x) is divided by (x-c1) (x-c2) ... (x-cm). At times Art. f(x)≥st. ((X-C1) (X-C2) ... (X-Cm)) = st. (x-c1) + art. (X-C2) + ... + Art. (x-cm) \u003d m, then st. f(x)≥m, and m is the number of roots of the rich term that can be considered. And the axis of the zero rich term is infinitely rich in roots, even if it has a meaning for whatever x is more beautiful 0. Zokrema, for the sake of causing it, and do not punish the same singing step. From well-proven theorems, the same assertion is evident.

If the polynomial f(x) is not a multinomial of step, greater, lower n, and may be greater, lower n roots, then f(x) is a zero polynomial. Indeed, from the mind of the mind of the firm, it is clear that f (x) is a zero polynomial, or art. f(x) ≤n. Assuming that the polynomial f(x) is not zero, then art. f(x) ≤n, and then f(x) can not be more, below n roots. We're getting to the point of superbness. Hence, f(x) is a non-zero rich term. Let f(x) and g(x) be non-zero rich terms of the step, not greater, lower n. If q polynomials acquire the same value for n + 1 values of change x, then f (x) = g (x).

For the proof, let's look at the rich term h(x) = f(x) – g(x). It dawned on me that - either h (x) = 0, or st. h (x) ≤n, then h (x) is not a rich term of the step, greater than, lower than n. Let me now take the number so that f (c) = g (c). Then h(c) = f(c) - g(c) = 0, then h is the root of the polynomial h(x). Also, the rich term h(x) has n+1 roots, and if, as it has been done, h(x) = 0, then f(x) = g(x). If f(x) and g(x) have the same values for all values of variable x, then

Multiple roots of the multinomial As the number є is the root of the multinomial f (x), this polynomial, apparently, is divisible by x-s. Might be possible that f(x) can be extended to the next step bugato-member x-s, i.e., on (x-c) k, k>1. This vipadka is called a multiple root. Let's formulate the appointment more clearly. The number is called the root of multiplicity k (k-fold root) of the polynomial f (x), so the polynomial is divisible by (x-c) k, k>1 (k is a natural number), but not divisible by (x-c) k + 1. If k=1, then it is called a simple root, and if k>1, it is called a multiple root of the polynomial f (x).

So the polynomial f(x) can be represented as f(x)=(x-c)mg(x), m is a natural number, vin is divisible by (x-c) m+1 and then if g(x) is divisible by x-c . Indeed, if g(x) is divisible by x-c, then g(x)=(x-c)s(x), then f(x)=(x-c) m+1 s(x), and also, f(x ) is subdivided by (x-c) m+1. Back, since f(x) is divisible by (x-c) m+1, then f(x)=(x-c) m+1 s(x). Then (x-c) mg (x) \u003d (x-c) m + 1 s (x) and after the short time for (x-c) m, g (x) = (x-c) s (x) is taken. It sounds like g(x) is subdivided into x-s.

It is clear, for example, that chi is the number 2 as the root of the rich term f(x) = x 5 -5 x 4+3 x 3+22 x 2 -44 x+24, and if so, then we know its multiplicity. In order to verify the first power supply, we can check for the additional Horner scheme, which divides f(x) by x-2. may be: Table 4. 2 1 1 -5 -3 3 -3 22 16 -44 -12 24 0 Like Bachimo, the excess when splitting f(x) by x-2 is more than 0, so it should be divided by x-2. Hence, the 2-root of the polynomial. In addition, we took away that f(x)=(x-2)(x 4 -3 x 3 -3 x 2+16 x-12). Now it’s obvious, chi є f (x) on (x-2) 2. Tse to deposit, how mi schoyno brought, in view of the divisibility of the polynomial g (x) \u003d x 4 -3 x 3 -3 x 2 + 16 x-12 on x-2.

Again speeding up by Horner's scheme: Table 5. 1 -3 -3 16 -12 2 1 -1 -5 6 0 -x2 -5x +6). Then f (x) \u003d (x-2) 2 (x 3 -x 2 -5 x + 6). Also, f(x) is divisible by (x-2) 2, now it is necessary to say that f(x) is divisible by (x-2)3. For which it is reversible that h(x) \u003d x 3 -x 2 -5 x + 6 is divided by x-2: Table 6. 1 -1 -5 6 2 1 1 -3 0 x-2, also, f(x) is divided by (x-2) 3, i f(x)=(x-2)3(x 2+x-3).

Then, similarly, it is possible to check whether f(x) is divided by (x-2)4, so that s(x)=x 2+x-3 is divided by x-2: Table 7. 2 1 1 1 3 -3 3 It is known that that the excess when s(x) is subdivided by x-2 is equal to 3, then s(x) is not subdivided by x-2. Also, f(x) does not subsume on (x-2) 4. In this way, f(x) subsumes on (x-2)3, but does not subsume on (x-2)4. Also, the number 2 is the root of the multiplicity of the rich term 3 f(x).

Sound the reverb of the root for the multiplicity of counting less at the table. For this application, the table may look like this: Table 8. 1 -5 3 22 -44 -24 2 2 1 1 -3 -1 1 3 -3 -5 -3 3 16 6 0 -12 0 0 Horner subtracted the multinomial f (x) by x-2, in another row we take away the coefficients of the polynomial g (x). Then let's take this other row into the first row of the new Horner system and subtract g (x) by x-2 and so on. In this way, the multiplicity of the root is equal to the number of otrimanih zero surpluses. In a row, to avenge the remaining non-zero surplus, there are also coefficients of the part when f (x) is subdivided by (x-2) 3.

Now, vikoristovuyuchi schoyno proponovan scheme of reverification of the root for the multiplicity, it seems that the task is coming. For any a and b, the rich term f(x) \u003d x 4 + 2 x 3 + ax 2 + (a + b) x + 2 can the number - 2 be the root of the multiplicity of 2? So the multiplicity of the root - 2 is due to add 2, then, having subdivided it by x + 2 for the proponated scheme, we are due to doubles to take the excess of 0, and in the third - the excess, which is equal to zero. May: Table 9. -2 -2 -2 1 1 2 0 -2 -4 aa a+4 a+12 a+b -3 a+b-8 2 2 a-2 b+2

In this rank, the number - 2 є root of the multiplicity of 2 of the expiratory rich term, then and only then, if

The rational root of the polynomial If the non-short term l/m (l, m are the integers of the number) is the root of the rich term f(x) with multiple coefficients, then the highest coefficient of the polynomial is divisible by m, and the long term is divisible by 1. True, like f (x )=anxn+an-1 xn-1+…+a 1 x+a 0, an≠ 0, de an, an-1, . . . , a 1, a 0 are integers, then f(l/m) = 0, then an(l/m) n+an-1 (l/m) n-1+. . . +a 1 l/m+a 0=0. Multiply the offending parts of the price of equivalence by mn. Take anln+an-1 ln-1 m+. . . +a 1 lmn-1+a 0 mn=0. Anln=m (-an-1 ln-1 -...- a 1 lmn-2 -a 0 mn-1) sounds.

Bachimo, the whole number anln is divisible by m. Ale l/m is a non-short drib, so the numbers l and m are mutually simple, but also, as per the theory of the validity of integer numbers, the numbers ln and m are also mutually simple. Otzhe, anln to be divided into m and m is mutually simple from ln, also, an to be divided into m. We know the rational root of the rich term f (x) \u003d 6 x 4 + 13 x 2 -24 x 2 -8 x + 8. According to the theorem, the rational root of the polynomial is found among non-short fractions in the form l / m, de l is the dilnik of the free term a 0 \u003d 8, and m is the dilnik of the highest coefficient a 4 \u003d 6. if so, then l / m is negative, then the sign "-" comes up to the number dial. For example, - (1/3) = (-1)/3. Also, we can say that l is the factor of the number 8, and m is the positive factor of the number 6.

The oscillators of the number 8 - tse ± 1, ± 2, ± 4, ± 8, and the positive dilators of the number 6 will be 1, 2, 3, 6, then the rational root of the looked rich term is among the numbers ± 1, ± 1/2, ± 1 /3, ±1/6, ±2/3, ±4, ±4/3, ±8/3. Guess we wrote down more than short fractions. In this order, we may have twenty numbers - "candidates" for roots. It was only left to reconsider the skin of them and choose those, as if true to the roots. A theorem is coming that will make it easier for the robot. As long as l/m is the root of the multiple term f(x) with multiple coefficients, then f(k) is divided by l-km for any integer k for mind, that l-km≠0.

To prove the theorem, we divide f(x) into x-k іz too much. We subtract f(x)=(x-k)s(x)+f(k). Oskіlki f(x) is a rich term with qlimi coefficients, then such a rich term is s(x), and f(k) is a whole number. Let s(x)=bn-1+bn-2+…+b 1 x+b 0. Then f(x)-f(k)=(x-k) (bnxn-1+bn-2 xn-2+ … +b1x+b0). Let's pay for this equanimity 1 x=l/m. If f(l/m)=0, then f(k)=((l/m)-k)(bn-1(l/m)n-1+bn-2(l/m)n- 2+…+b 1(l/m)+b 0). Multiply the offending part of the remaining equity by mn: mnf(k)=(l-km)(bn-1 ln-1+bn-2 ln-2 m+…+b 1 lmn-2+b 0 mn-1). It is clear that the number mnf (k) is divided by l-km. Ale oskіlki l і m are mutually simple, then mn і l-km are also mutually simple, also, f (k) is divided by l-km. The theorem has been completed.

Let's turn to our butt and, after having proved the theorem, it is even more sonorous about the sound of the rational root. It is necessary to assign the theorem for k=1 і k=-1, that is, because the non-short drіb l/m is the root of the term f(x), then f(1)/(l-m), and f(-1)/(l + m) . It is easy to know that in times f(1)=-5, and f(-1)=-15. Respectfully, at the same time, we turned it off at a glance ± 1. Henceforth, the rational root of our rich term is the following number of middle numbers ± 1/2, ± 1/3, ± 1/6, ± 2, ± 2/3, ± 4/3, ± 8 /3. Let's look at l/m=1/2. Then l-m=-1 and f(1)=-5 are divided by the whole number. Dalі, l+m=3 і f(1) =-15 so itself is divided by 3. So, drіb 1/2 is left in the middle of "candidates" at the root.

Let me now lm=-(1/2)=(-1)/2. In this case, l-m=-3 і f(1) =-5 does not divide by - 3. So, drіb -1/2 cannot be the root of this rich term, and we can turn it off from a distant view. It is necessary to reconsider for skin s in writing shots, we take into account that the root is found among the numbers 1/2, ± 2/3, 2, - 4. In this rank, to finish the same simple trick, the region of the rational roots of the considered polynomial sounded meaningful. Well, for rechecking the numbers that are left out, we can use the Horner scheme: Table 10. 6 13 -24 -8 8 1/2 6 16 -16 0

Bachimo, scho 1/2 is the root of the rich term f(x) and f(x) = (x-1/2) (6 x 3+16 x 2 -16 x-16) = (2 x-1) (3 x 3+8 x 2 -8 x-8). It was clear that the other roots of the polynomial f(x) are taken from the roots of the polynomial g(x) =3 x 3+8 x 2 -8 x-8, then, further rechecking of the "candidates" in the root can be carried out already of the same polynomial. We know: Table 11. 3 8 -8 -8 2/3 3 10 -4/3 -80/9 We took away that the excess when g(x) was subdivided by x-2/3 is more - 80/9, then. 2/3 is not a root of the polynomial g(x), also, i f(x). Further, we know that - 2/3 is the root of the polynomial g (x) and g (x) \u003d (3 x + 2) (x 2 + 2 x-4).

Then f(x) = (2x-1) (3x+2) (x2+2x-4). Further verification can be carried out for the polynomial x 2+2 x-4, which is remarkably simpler, lower for g (x) or larger for f (x). As a result, it is taken into account that the numbers 2 i - 4 are not rooted. Also, rich term f(x) =6 x 4+13 x 3 -24 x 2 -8 x+8 has two rational roots: 1/2 i - 2/3. This method makes it possible to know only a rational root of a rich term with a large number of coefficients. Tim is sometimes a rich member of the mother and irrational root. So, for example, when looking at the butt of a rich term, there are only two roots: - 1±√5 (these root of a rich term is x2 + 2 x-4). a polynomial can be called a non-material rational root.

When testing "candidates" at the root of the rich term f(x) after further elaboration of other theorems, you should call the left for the candidates k=± 1. In other words, if l/m is a "candidate" at the root, then you will overthink that f( 1 ) and f(-1) on l-m and l+m are correct. But it could be, for example, f(1) =0, i.e. 1 is the root, then f(1) can be extended as a number, and the re-verification takes sense. In this case, divide f(x) by x-1, so take f(x)=(x-1)s(x) and test for the polynomial s(x). If you forget that one root of the polynomial f(x)-x 1=1 - we already knew. If the "candidates" are reversed at the root, which have been lost after another theorem about the rational root, after Horner's scheme it is possible that, for example, l / m is the root, then you should know its multiplicity. If it is more expensive, say, k, then f(x)=(x-l/m) ks(x), and further reverification can be done for s(x), which will shorten the calculation.

Solution. After changing the change y=2 x, let's move on to a polynomial with a coefficient equal to one for the highest step. For this shoulder, we multiply the viraz by 4. If the function of the root is taken away, then the stench is found in the middle of the free member. Writeable ix: ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15±, ±20, ±30, ±60

We calculate sequentially the value of the function g(y) at these points up to zero. Tobto, y=-5 є root, otzhe, є root of external function. Conducted under the stovpchik (coil) of the rich term on the binomial

The re-verification of dilnikov, which is lost, should be carried out incompletely, so it’s easier to lay out the square trinomial Otzhe into multipliers of subtractions,

Vykoristannya formulas of fast multiplication and Newton's binomial for the expansion of a rich term into factors Inodi old look polynomial to suggest about the method of spreading yoga on multipliers. For example, after inconsistent transformations, the coefficients of vishikovyvayutsya in a row from Pascal's tricot for the coefficients of Newton's binomial. butt. Lay out the multiplier term.

Solution. We turn it around to the point: The sequence of coefficients in sum in the arms clearly shows what it is. From the same, Now, we will formulate the formula for the difference of squares: Viraz the other arc does not have action roots, but for the rich term from the first arc, we once again formulate the formula for the difference of squares

Vieta's formulas express the coefficients of a polynomial through the th root. With these formulas, you can manually correct the correctness of the meaning of the root of the rich term, as well as for the folding of the rich term for given roots. The formula As a root of a polynomial, then the coefficients are manifested by the symmetrical rich terms of the roots, and

In other words, ak dear sum of all possible creations from k roots. As the senior coefficient of the polynomial, then it is necessary to divide all the coefficients into a 0 in advance of the Vieta formula. From the rest of the formula Vієta is strong, as if the root of the rich member is integer, then the stench is the dilniks of the yogo free member, which is also integer. The proof is based on the view of equivalence, taking away the arrangement of the rich term according to the roots, vrakhovuchi, scho a 0 = 1 Equating the coefficients at the same steps x is obsessed with the formula Vієta.

Untie the alignment x 6 – 5 x 3 + 4 = 0 Untie. Significantly y \u003d x 3, even though it is equal to look at y 2 - 5 y + 4 \u003d 0, otherwise Y 1 \u003d 1; Y 2 \u003d 4. Otzhe, vyhіdne r_vnyannya is equivalent to the marriage of rіvnyan: x 3 \u003d 1 chi x 3 \u003d 4, i.e. X 1 \u003d 1 chi X 2 \u003d Vidpovіd: 1;

Bezout Destination Theorem 1. An element is called the root of a rich term, so f(c)=0. Bezout's theorem. The excess in the subdivision of the polynomial Pn(x) by the binomial (x-a) increases the value of the polynomial at x = a. Bringing. By virtue of the algorithm, f(x)=(xc)q(x)+r(x), de or r(x)=0, otherwise. Later, f(x)=(x-c)q(x)+r, later, f(c)=(c-c)q(c)+r=r, and f(x)=(xc)q(x) +f(c).

Last 1: The excess in the subdivision of the polynomial Pn (x) by the binomial ax+b is more valuable for the polynomial at x = -b/a, then R = Pn (-b/a). Last 2: As the number a is the root of the polynomial P (x), whose polynomial is divisible by (x-a) without excess. Lesson 3: How the polynomial P(x) can be pairwise different roots a 1 , a 2 , … , an, vin dividing by tvir (x-a 1) … (x-an) without excess. Lesson 4: A rich member of the step n may be three or more more than n different roots. Lesson 5: For any polynomial P(x) that number a is different (P(x)-P(a)) divisible without excess by the binomial (x-a). Lesson 6: The number a is the root of the polynomial P(x) of degree not lower than the first and only if P(x) is divided by (x-a) without excess.

The layout of a rational fraction in the simplest way Let it be given the correct rational argument (1).

Theorem 1. Let x=а є the root of the banner of style k, then , de f(a)≠ 0, then the same correct fraction can be given at the sum of two other regular fractions in the coming order: (2) , and F 1 (x) is a rich term, the step of which is lower than the step of the standard

de richomember, the step of some kind of lower step of the standard. І similarly to the forward formula can be taken: (5)

As we have already designated, one of the most important tasks of the theory of richly defined terms is the task of understanding their roots. For the accomplishment of this task, you can win the selection method, tobto. take a real number and change it, which are the roots of this polynomial.

With this, you can drink shvidko on the root, or you can not know it at all. It is impossible for aje to pervert all the numbers, for those who are too rich.

In addition, we managed to sound the region by a joke, for example, the nobility, which is rooted, say, in the middle of thirty specified numbers. And for thirty numbers, you can also work on a reverb. At the link with the mustache, we say more importantly, and we see such a firmness.

As long as l/m (l,m - integers of the number) is the root of the multiple term f(x) with the integer coefficients, then the higher coefficient of the polynomial is divisible by m, and the larger term is divisible by 1.

Indeed, if f(x) = anxn+an-1xn-1+... +a1x+a0, an?0, de an, an-1,...,a1, a0 are integers of a number, then f (l /m) = 0, then an (l/m) n+an-1 (l/m) n-1+...+a1l/m+a0=0.

Multiply the offending parts of the price of equivalence by mn. Take anln+an-1ln-1m+... +a1lmn-1+a0mn=0.

Sounds are screaming:

anln=m (-an-1ln-1-... - a1lmn-2-a0mn-1).

Bachimo, the whole number anln is divisible by m. Ale l / m - not a short drіb, tobto. the numbers l and m are mutually simple, also, as per the theory of the divisibility of integers, the numbers ln and m are also mutually simple. Otzhe, anln to be divided into m and m is mutually simple from ln, also, an to be divided into m.

The topic has been brought up to allow the area to be meaningfully sounded by the search for a rational root of a rich term with multiple coefficients. We will demonstrate it on a specific application. We know the rational root of the rich term f(x) = 6x4+13x2-24x2-8x+8. According to the theorem, the rational root of the polynomial is found in the middle of non-short fractions in the form l / m, de l is the dilnik of the long term a0 = 8, and m is the dilnik of the highest coefficient a4 = 6. if so, yakscho drіb l/m is negative, then the sign "-" vodnosimeme to the numeral. For example, - (1/3) = (-1)/3. Also, we can say that l is the factor of the number 8, and m is the positive factor of the number 6.

The oscillators of the number 8 - tse ±1, ±2, ±4, ±8, and the positive dilators of the number 6 will be 1, 2, 3, 6, then the rational root of the examined rich term is the middle of the numbers ±1, ±1/2, ± 1 /3, ±1/6, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3. Guess we wrote down more than short fractions.

In this order, we may have twenty numbers - "candidates" for roots. It was only left to reconsider the skin of them and choose those, as if true to the roots. But again, I’ll have to do a lot of reworking. And the axis is coming, the theorem will make it easier for the robot.

As long as l/m is the root of the multiple term f(x) with multiple coefficients, then f(k) is divided by l-km for whatever integer k is, for example, l-km?0.

To prove the theorem, we divide f(x) into x-k іz too much. Take f (x) = (x-k) s (x) +f (k). Since f(x) is a rich term with multiple coefficients, then such a polynomial is s(x), and f(k) is a whole number. Let s(x) = bn-1+bn-2+…+b1x+b0. Then f(x) - f(k) = (x-k) (bn-1xn-1+bn-2xn-2+...+b1x+b0). Let's pay for this equanimity x=l/m. Vrahovoyuchi, scho f (l / m) = 0, it is possible

f (k) = ((l/m) - k) (bn-1 (l/m) n-1+bn-2 (l/m) n-2+…+b1 (l/m) +b0) .

Multiply the offending part of the remaining jealousy by mn:

mnf (k) = (l-km) (bn-1ln-1+bn-2ln-2m+…+b1lmn-2+b0mn-1).

It is clear that the number mnf (k) is divided by l-km. Ale oskіlki l і m are mutually simple, then mn і l-km are also mutually simple, also, f (k) is divided by l-km. The theorem has been completed.

Let's turn now to our butt and, after having proved the theorem, it sounds even more loudly when it comes to the sound of the rational root. It is necessary to assign the theorem for k=1 і k=-1, so. as a non-short drіb l/m is the root of f(x), then f(1)/(l-m), and f(-1)/(l+m). It is easy to know that f(1) =-5, and f(-1) =-15. Respectfully, we turned off the contagion at a glance ±1.

Also, the rational root of our rich term is the following of the middle numbers ±1/2, ±1/3, ±1/6, ±2, ±2/3, ±4, ±4/3, ±8, ±8/3.

Let's look at l/m=1/2. Then l-m=-1 and f(1)=-5 are divided by the whole number. Dalі, l+m=3 і f(1) =-15 so itself is divided by 3. So, drіb 1/2 is left in the middle of "candidates" at the root.

In this rank, to finish the same simple trick, they meaningfully sounded the region in search of a rational root of the analyzed polynomial. Well, for re-checking the numbers, we use Horner's scheme:

Table 10

They took away that the excess when g (x) was subdivided by x-2/3 is equal to 80/9, so 2/3 is not the root of the rich term g (x), but means, i f (x).

Further, it is easy to know that - 2/3 is the root of the multiple term g(x) and g(x) = (3x+2) (x2+2x-4). Then f(x) = (2x-1) (3x+2) (x2+2x-4). Further verification can be carried out for the polynomial x2+2x-4, which is obviously simpler, lower g(x) or larger f(x). As a result, it is taken into account that the numbers 2 i - 4 are not rooted.

Also, the rich term f(x) = 6x4+13x3-24x2-8x+8 has two rational roots: 1/2 i - 2/3.

Guessing, more descriptions of the method gives the possibility to know the rational root of the rich term with many coefficients. Tim is sometimes a rich member of the mother and irrational root. So, for example, when looking at the butt of a rich member, there are only two roots: - 1±v5 (these root of a rich member is x2 + 2x-4). And, apparently, a rich member may not be the mother of a rational root.

Now the lady is happy.

When trying out "candidates" at the root of the rich term f(x), after further elaborating more theorems, sound the left for vipadkіv k=±1. In other words, since l/m is a "candidate" at the root, it is reversed whether f (1) and f (-1) can be divided into l-m and l+m obviously. But it could be that, for example, f (1) = 0, then 1 is the root, and then f (1) can be divided by a number, and our reverification takes sense. І here the next step is to divide f (x) by x-1, so. take f(x) = (x-1) s(x), and test for the polynomial s(x). If you don’t forget that one root of the rich term f(x) – x1=1 – we already knew. As in the case of reversing the "candidates" at the root, which was lost after another theorem about the rational root, Horner's scheme is taken away, for example, l / m - the root, then you should know its multiplicity. If it’s more expensive, let’s say, k, then f(x) = (x-l/m) ks(x), and further reverification can be done for s(x), which will shorten the calculation.

In this rank, we learned to know the rational root of the rich term with large coefficients. It appears that we ourselves learned to know the irrational root of the rich term with rational coefficients. In fact, as far as I can, for example, a rich term f (x) \u003d x4 + 2 / 3x3 + 5 / 6x2 + 3 / 8x + 2, then, having added the coefficients to the sleeping banner and adding yoga by the arms, we take f (x) \u003d 1 /24 (24x4+16x3-20x2+9x+48). It was clear that the roots of the polynomial f(x) are formed from the roots of the rich term, which stand at the arms, and in the new coefficient - the numbers. Let's say, for example, that sin100 is an irrational number. Speeding up with the home formula sin3?=3sin?-4sin3?. Stars sin300 = 3sin100-4sin3100. Looking back at those that sin300=0.5 and conducting awkward transformations, we can assume 8sin3100-6sin100+1=0. Also, sin100 is the root of the term f(x) = 8x3-6x+1. Just as we shukatimemo rationally the root of that rich member, then we perekaєmosya, we don’t have them. Otzhe, the root of sin100 is a rational number, tobto. sin100 is an irrational number.

Come on

- rich term of step n ≥ 1 in the effective value of the complex variable z with the effective value of the complex coefficients a i . Let's prove the following theorem.

Theorem 1

Leveling P n (z) = 0 May I want one root.

Let's come Lema.

Lemma 1

Let P n (z)- rich term of step n, z 1 - the root of the river:
P n (z1) = 0.
Todi P n (z) can be revealed in one way by looking at:
P n (z) = (z - z 1) P n-1 (z),
de P n- 1(z)- rich term step n - 1 .

Bringing

To prove it, we will make a theorem (div. The division of a multiple of a multiple term by a multiple of a multiple term by a fold and a stump), it is possible for any two multiple terms P n (z) i Qk (z), steps n and k, moreover, n ≥ k
P n (z) = P n-k (z) Q k (z) + U k-1 (z),
de P n-k (z)- rich term of step n-k, U k- 1(z)- rich term of the step is not higher than k- 1 .

Let's put k = 1 , Qk (z) = z - z 1 also
P n (z) = (z - z 1) P n-1 (z) + c,
de c - fast. Imagine here z = z 1 that vrahuєmo, scho P n (z1) = 0:
P n (z 1 ) = (z 1 - z 1 ) P n-1 (z 1 ) + c;
0 = 0 + c.
Zvіdsi c = 0 . Todi
P n ,
what it was necessary to bring.

The expansion of the rich term into multipliers

Also, on the basis of Theorem 1, the rich term P n (z) May I want one root. Significantly yogo yak z 1 , P n (z1) = 0. Same on stand lemy 1:
P n (z) = (z - z 1) P n-1 (z).
Dali, like n > 1 , then the polynomial P n- 1(z) so can I want one root, which is meaningful like z 2 , Pn- 1(z2) = 0. Todi
Pn- 1 (z) = (z - z 2) P n-2 (z);
P n (z) = (z - z 1) (z - z 2) P n-2 (z).

Continuing this process, we come to the conclusion that we have n numbers z 1, z 2, ..., z n such that
P n (z) = (z - z 1) (z - z 2) ... (z - z n) P 0 (z).
Ale P 0 (z)- tse postiyna. Equating the coefficients at z n , it is known that it is more expensive a n . As a result, we are obsessed with the formula for dividing a rich term into multipliers:
(1) P n (z) = a n (z - z 1) (z - z 2) ... (z - z n).

Numbers z i є to the roots of the rich term P n (z).

At the zagalny vpadku not all z i, scho to enter before (1) , Riznі. Among them there can be the same values. How to expand a rich term into multipliers (1) you can write at the sight:
(2) P n (z) = a n (z - z 1 ) n 1 (z - z 2 ) n 2 ... (z - z k ) n k;
.
Here z i ≠ z j for i ≠ j. Yakscho n i = 1 , then root z i called forgive. Vіn enter at the layout for multipliers at the sight (z-z i ). Yakscho n i > 1 , then root z i called the multiple root of the multiplicity n i . Vіn enter at the layout of the multipliers when looking at the extraction of n i prime multipliers: (z-z i )(z-z i ) ... (z-z i ) = (z-z i ) n i.

Rich terms with effective coefficients

Lemma 2

Since it is a complex root of a polynomial with effective coefficients, then the number is also complexly related to the root of the polynomial, .

Bringing

Deisno, yakscho, and polynomial coefficients - dіysnі numbers, then.

In this order, the complex root is included in the layout on the multipliers in pairs with their complex meanings:
,
de, - Real numbers.
Same layout (2) a rich term with effective coefficients for multipliers can be filed at the sight, in the presence of only effective fast:
(3) ;
.

Methods for splitting a rich term into multipliers

With the improvement of what was said above, for the decomposition of a polynomial into factors, it is necessary to know all the roots of the equation P n (z) = 0 and designate their multiplicity. Multipliers with complex roots need to be grouped in a complex way. Same layout depends on the formula (3) .

In this rank, the method of spreading the rich term into multipliers is used in the offensive:
1. We know root z 1 equalization P n (z1) = 0.
2.1. Yakshcho root z 1 effective, then in the layout we add a multiplier (z-z1) (z-z1) 1 :
.
1(z), starting from point (1) , Until we know all the roots.
2.2. As a complex root, the number є is complexly obtained as the root of a rich term. Todі before laying out enter the multiplier

,
de b 1 = - 2 x 1, c 1 = x 1 2 + y 1 2.
In my mind, in the layout we add a multiplier (z 2 + b 1 z + c 1) i dilute the rich term P n (z) by (z 2 + b 1 z + c 1). As a result, we take a rich term of step n - 2 :
.
Let's repeat the process for the polynomial P n- 2(z), starting from point (1) , Until we know all the roots.

Knowledge of the root of the rich member

head office, with the expansion of the polynomial into factors, the significance of the yogo root. Unfortunately, you can’t always work analytically. Here we will analyze the sprat of vipadkiv, if you can know the root of the rich term analytically.

Root of the rich member of the first stage

The rich member of the first step is an integral function. There is only one root. The layout can only be one multiplier to avenge the change of z:
.

Root of a rich member of another level

In order to know the root of the rich term of another level, it is necessary to untie the square equal:
P 2(z) = a 2 z 2 + a 1 z + a 0 = 0.
As a discriminant, then there are two real roots:
, .
Just look at the multipliers:
.
What is the discriminant D = 0 , then equal may one dvorazovy root:
;
.
As discriminant D< 0 , then the root is more complex,
.

Richly articulated step higher for another

Іsnuyu formulas for the meaning of the roots of the rich segments of the 3rd and 4th steps. They rarely cory with them, the shards of the stench are bulky. There are no formulas for the knowledge of the roots of the richly articulated degree higher than the 4th. Ignorantly on the spot, in deyakih vipadkas, one goes into spreading the rich term into multipliers.

Significance of the whole root

It seems that it is a rich term, for some coefficients - the number of numbers, the number of roots, which can be known by sorting through all the possible values.

Lemma 3

Give me a rich dick
,
coefficients a i of which - the number of the number that can be the root of z 1 . Same root as dilnik of number a 0 .

Bringing

Let's rewrite equal P n (z1) = 0 at the sight:
.
Todi - tsile,
Mz 1 = - a0.
Divided by z 1 :
.
Oskіlki M - qile, then i - qile. What did it take to bring.

Therefore, as the coefficients of a polynomial - the numbers of numbers, you can try to know the numbers of the root. For whom it is necessary to know all the dilniks of a free member 0 і, equalization substitution P n (z) = 0, perverti, chi є stink to the roots of that equal.
Note. Since the coefficients of a polynomial are rational numbers, then multiplying equal P n (z) = 0 on the high standard of the numbers a i we take the equalization for the polynomial with integer coefficients.

The meaning of the rational root

Since the coefficients of a polynomial - the numbers of the number and the number of roots are not, then for a n ≠ 1 , you can try to know the rational root. For whom is it necessary to create a substitution
z = y/a n
and multiply equal by a n n- 1 . As a result, we take into account the equality for the rich term in the form of change and with the number of coefficients. Dali shukaimo the root of the rich member of the middle member of the free member. Since we knew such a root y i , then passing to the change x , we will assume a rational root
z i = y i / a n.

Colored formulas

We introduce formulas, with the help of which it is possible to expand the polynomial into factors.

Have a more savage temper, to lay out a rich member
P n (z) = z n - a 0,
de a 0 - it is more complex, it is necessary to know all the yogo roots, so that you can unravel equal:
z n = a 0 .
Tsіvnyannya is easy to be mistaken, as if to prove a 0 via module r i argument?
.
Oskilki a 0 do not change, as to add to the argument 2 π, then imagine a 0 at the sight:
,
de k – qile. Todi
;
.
Assigning values k k = 0, 1, 2, ... n-1, We take n roots of the polynomial. Todi yogo layout for multipliers may look:
.

Bisquare bagatonic term

Let's take a look at the biquadratic term:
.
A biquadrate rich term can be divided into multipliers, without a root.

When , maybe:

,
de.

Bicubic and rich segments that can be reduced to a square

Let's look at the rich member:
.
Yogo root stands for equal:
.
Won be guided up to square alignment substitution t = z n :
a 2 n t 2 + a n t + a 0 = 0.
Virishivshi tse eve, we know the yogo root, t 1 , t 2 . If we know the arrangement at the sight:
.
Dali by the method, let's look at it, expand it into multipliers z n - t 1 i z n - t 2 . The visnovka has a group of multipliers, which avenge the root in a complex way.

Rotary stalks

The rich member is called return yakscho yogo coefficients are symmetrical:

Butt of the storable bagato-member:
.

Since the steps of the reverse polynomial n are unpaired, such a polynomial can have root z = -1 . Dividing such a rich term into z + 1 , we take the return rich term of the step

In case of rozv'yazannі rivnyan i nerіvnjanі often vykaє nіkaє nebhіdnіst razvіdnіє polynomial razvіdnіє on polynomials, stupіnіy і і dіvnіє three or more. We can look at these stats, how to make it simpler.

Like a zavzhd, beastly for help to theory.

Bezout's theorem stverzhuє, scho surplus in splitting a polynomial into a binomial dorivnyuє.

But what is important for us is not the theorem itself, but consequence from it:

Since the number is the root of a polynomial, then the polynomial can be divided without too much binomial.

Before us is the task of knowing how to know one root of a rich term, then we divide the rich term into, de - the root of a rich term. As a result, we take a rich member, the foot of one is one lesser, the lower is the rib of the outer one. And then for consumption, you can repeat the process.

Tse zavdannya split into two: how to know the root of a rich term, and how to divide a rich term into a binomial.

Let us report on these moments.

1. How to know the root of a rich member.

The back of the hand is revered, chi is the number 1 and -1 of the roots of the rich member.

Here are some facts to help us:

As the sum of all coefficients of the polynomial is equal to zero, the number is the root of the polynomial.

For example, the polynomial of the sum of the coefficients is equal to zero: . It is easy to misinterpret what is the root of a rich member.

As the sum of the coefficients of the polynomial at paired steps is the same as the sum of the coefficients at unpaired steps, the number is the root of the polynomial. Vilniy member vvazhaetsya coefficient at the double level, oskolki, and - guy number.

For example, in the polynomial of the sum of coefficients at paired steps : , and the sum of coefficients at unpaired steps : . It is easy to misinterpret what is the root of a rich member.

If nі 1, nі -1 є to the roots of the polynomial, then the distance collapses.

For the induced rich term of the step (tobto the rich term, in which the senior coefficient is the coefficient at - the leading one), the following formula is valid:

De - the root of the rich member.

There are more formulas of Vієta, that there are other coefficients of the polynomial, but we can talk about it ourselves.

Z tsієї formula Vієta viplivaє, scho as the root of a rich member of the integer, then the stench of the dilniks of the yogo free member, which is also a whole number.

Vihodyachi z tsogo, we need to lay out the variable term of the rich term into multiples, and sequentially, from the smallest to the largest, reverse, which of the plurals is the root of the rich term.

Look at it, for example, rich member

Free member's diaries: ; ; ;

The sum of all coefficients of a polynomial is more expensive, then, the number 1 has ceased to be the root of a polynomial.

The sum of the coefficients for the twin steps:

Sum of coefficients for unpaired steps:

Also, the number -1 is also the root of a polynomial.

It is reversible that chi is the number 2 as the root of a rich term: also, the number 2 is the root of a rich term. Later, following Bezout's theorem, a rich term can be divided without excess into a binomial.

2. How to subtract a rich term into a binomial.

The rich term can be divided into a binomial with a stump.

We divide the rich term into a binomial with a stompchik:

The second way to subdivide a polynomial into a binomial is Horner's scheme.

Watch the video to understand how to divide a rich term into a binary term with a step i for the additional Horner's scheme.

I’ll respect that when rozpodіlі stovpchik like steps unfamiliar to the vyhіdny polynomial vіdsutnya, її mіstsі write 0 - like і, like from the folded table for Horner’s scheme.

Also, as we need to divide the rich term into a binary term and as a result we take the rich term, then we can know the coefficients behind Horner's scheme:

We can also vicorist Horner's scheme in order to reverse, chi is given the number of the root of the rich term: if the number is the root of the rich term, then the excess in the subfield of the rich term is equal to zero, so in the remaining column of the other row of the Horner scheme we take 0.

Vikoristovuyuchi Horner's scheme, we "drive in two birds with one stone": one hour it is reverified, chi is the number of the root of a rich term and dividing a rich term into a binomial.

butt. Virishiti Rivnyannia:

1. Write down the dilniks of the free member, and shukatimemo the root of the rich member of the middle dilniks of the free member.

Dialogues of the number 24:

2. Reversibly, chi is the number 1 root of a rich term.

The sum of the coefficients of the polynomial, also, the number 1 is the root of the polynomial.

3. Divide the outward rich term into a binary term using Horner's scheme.

A) Write down the first row of the table of coefficients of the output polynomial.

Oskіlki member, scho vengeance vіdsutnya, at that table table, which may have a coefficient when we write 0. We write evil root of knowledge: number 1.

B) Save the first row of the table.

In the rest of the column, as if it were clear, we subtracted zero, the world divided the last rich term into a binomial without excess. The coefficients of the polynomial, which the result has under the image in blue color in another row of the table:

It is easy to misunderstand that the numbers 1 and -1 are not the roots of a rich term

C) We continue the table. Reversibly, chi is the number 2 as the root of a rich term:

So the step of the polynomial, which appears in the result of the sub-term is one less than the step of the output rich term, also the number of coefficients and the number of columns are one less.

In the rest of the column, we took away -40 - a number that does not add up to zero, therefore, the rich term is divided by a binary term from the excess, and the number 2 is not the root of the rich term.

C) Reversibly, chi is the number -2 as the root of a rich term. So, as before, the test was not far away, so that there was no swindle with coefficients, I’m in a row, that I’m confirming my test:

Miraculous! Zero was taken away from the excess, then, the rich term was divided into a binomial without excess, and the number -2 is the root of the rich term. The coefficients of the polynomial, which in the result subdivides the polynomial into a binomial in the table of the image in green color.

As a result, we subtracted the square trinomial , the root of which is easy to know behind the Viet theorem:

Otzhe, the root of the outward revival:

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Suggestion: ( }

Yakscho rich member

Bringing

Let us have the coefficients of the polynomial є by integer numbers, and let the number a be the root of the th rich term. To the one in which the sound shines in every moment, the coefficient is divided by a.

Respect. This theorem actually allows you to know the roots of the richest terms in that case, if the coefficients of these rich terms are numbers, and the root is rational number. The theorem can be reformulated as follows: just as we know that the coefficients of a polynomial are the numbers of the number, and the root of the yogo is rational, then the rational root can only be like de p as a dilnik of a number (a free term), and the number q is a dilator of a number (a senior coy) .

The theorem about the whole root, what to take revenge on yourself

Just as the number α is the root of a rich term with multiple coefficients, then α is the dilnik of the yogic free term.

Bringing. Come on:

P(x)=a 0 xⁿ +a 1 xⁿ -1 +…+a n-1 x +a n

rich term with qlimi coefficients and qile number α - yogo root.

Then the value of the root is equalized P (α)=0;

a 0 αⁿ+a 1 αⁿ -1 +…+a n-1 α +a n =0.

Vinosyachi zagalny multiplier α for the bows, take away the equivalence:

α(a 0 αⁿ -1 +a 1 αⁿ -2 +…+a n-1)+a n =0 , stars

a n = -α(a 0 αⁿ -1 +a 1 αⁿ -2 +…+a n-1)

Shards of the number a 0 , a 1 ,…a n-1 , an i α −tsіlі, then the arcs should be the whole number, then, a n be divided by α, as it was supposed to be completed.

The theorem has been brought up, but it can be formulated in such a way: the number of roots of the polynomial with the number of coefficients is the dilator of the first free term.
On the theorem of foundations, the algorithm for searching for the integer root of a rich term with the whole number of coefficients:

2. Dodatkova theorem about the root value

In addition to the number of α-roots of the rich term P(x) with integer coefficients, then α-1-divisor of the number P(1), α+1-divisor of the number P(-1)

Bringing. 3 the sameness

xⁿ-yⁿ=(x-y)(xⁿ -1 +xⁿ -2 y+…+ xyⁿ -2 +yⁿ -1)

you can see that from the number of numbers b і c the number bⁿ-cⁿ is divisible by b∙c. Ale for any rich member P retail

P (b)-P(c)= (a 0 b+a 1 bⁿ -1 +…+a n-1 b+a n)-(a 0 cⁿ+a 1 cⁿ -1 +…+a n-1 c +a n)=

=a 0 (bⁿ- cⁿ)+a 1 (bⁿ -1 -cⁿ -1)+…+a n-1 (b-c)

і, also, for a polynomial P with zіlimi coefficients і zіlih numbers b і c the difference P(b)-P(c) is subdivided into b-c.

Let's remember: for b = α, z = 1, P (α)-P (1) = -P (1), which means that P (1) is divided by α-1. Similarly, there is another view.

Horner's scheme

Theorem: Let a short-term drіb p / q є root equal a 0 x n +a 1 x n − 1 + +a n − 1 x+a n =0 with multiple coefficients, the same number q є dilnik of the senior coefficient a0, and the number R є dilnik free member an.

Respect 1. Be the root of the relationship with the number of coefficients and dilnik of the yogo free member.

Respect 2.As the senior coefficient is equal to the number of coefficients of the road 1, all rational roots, as the stink is known - the number.

The root of the rich member. The root of a rich member f(x)= a 0 x n +a 1 x n − 1 + +a n − 1 x+a n є x = c , so what f (c)=0 .

Note 3. Yakscho x = c root of a rich member , then the rich term can be written as: f(x)=(x−c)q(x) , de tse private view under the rich member f(x) into a monomial x-c

You can subdivide a rich term into a monomial using Horner's scheme:

Yakscho f(x)=a 0 x n +a 1 x n − 1 + +a n − 1 x+a n , a0 ≠0 , g(x)=x−c , then when rozpodіlі f (x) on the g (x) privately q(x) may look q(x)=b 0 x n − 1 +b 1 x n − 2 + +b n−2 x+b n−1 , de b 0 = a 0 ,

b k = c b k − 1 + a k , k=1, 2, ,n−1. surplus r know the formula r=c b n − 1 +a n

Solution: The coefficient at the senior level is equal to 1; 2; 3; four; 6; 12. Vikoristovuyuchi Horner's scheme, we know the number of roots equal:

There is one root of choice for Horner's scheme. then you can do it like this x 3 −x 2 −8x+12=(x−2)(x 2 +x−6)=0 (x−2) 2 (x−3)=0 x=2;x=3