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  • We begin by [[equate | equating]] the two expressions: ...th>a</math>, <math>b</math>, and <math>c</math> are integers, we can match coefficients:
    3 KB (439 words) - 17:24, 10 March 2015
  • Equating coefficients, we have <math>2C - 64 = -(1 + 4 + \ldots + 225) = -1240</math>, so <math>- ...<math>\{2, 4, \ldots 14\}</math>. The sum of the negative <math>x^2</math> coefficients is the sum of the products of the elements in all two element sets such tha
    6 KB (941 words) - 10:37, 27 May 2024
  • ...3^{\circ}}</cmath> given the conditions of the problem. Equating imaginary coefficients, we have that <cmath>b \left( 1 - \frac{1}{a^2+b^2}\right) = 0</cmath> givi We now know that <math>a^2+b^2=1</math>, so when we equate real coefficients we have that <math>2a = 2 \cos{3^{\circ}}</math>, therefore <math>a = \cos{
    4 KB (671 words) - 19:24, 2 November 2024
  • ...+ (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)</math>. Equating coefficients, we get the following system of equations:
    4 KB (728 words) - 11:07, 2 December 2024
  • By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation, By equating the imaginary terms on each side of the equation,
    5 KB (850 words) - 11:53, 23 December 2023
  • Equating the coefficients of <math>x^3</math> and <math>x</math> with their known values, we are left ...(a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.</cmath>Matching coefficients, we get
    4 KB (704 words) - 18:25, 28 March 2024
  • ...d <math>q(x)</math> are both equal to <math>a</math>, and equating the two coefficients gives .../math> into <math>q(x)</math> yields another long polynomial. Equating the coefficients of <math>x</math> in both polynomials, we get:
    5 KB (972 words) - 18:10, 10 December 2024
  • ..., the coefficients of like terms on both sides must be equal. Equating the coefficients of <math>y</math> and the constant terms, we get the system of equations: a+b &= 2 \quad \text{(from coefficients of } y \text{)} \
    1 KB (212 words) - 18:53, 28 September 2023
  • Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying ...- b)(x - c) - (x - d)(x - e)(x - f)]</math>, so comparing <math>x^2</math> coefficients gives <math>1 + b + c = d + e + f</math> and thus both values equal to <mat
    8 KB (1,474 words) - 10:02, 15 August 2024
  • ...t be expressed as the product of two non-constant polynomials with integer coefficients. ...g up the chain until we get that <math>3|b_{n-2}</math>. Then, by equating coefficients once more, we get that <math>b_0c_{n-1}+b_1c_{n-2}+\ldots+b_{n-2}c_1+b_{n-1
    3 KB (530 words) - 06:08, 22 June 2024
  • ...is equal to the original equation <math>x^3-ax^2+bx-a</math>, equating the coefficients, we get that
    3 KB (584 words) - 17:36, 16 October 2024
  • Comparing coefficients with <math>f(x)</math>, we see that To find <math>f(1)</math> we just need to find the sum of the coefficients which is <math>1 + 1 - 8009 + 100 + 900= \boxed{\bold{(C)} \ , -7007}.</mat
    10 KB (1,726 words) - 19:55, 11 October 2024
  • for some number <math>k</math>. By comparing <math>x^4</math> coefficients, we see that <math>C=1</math>. Thus, Expanding and equating coefficients we get that
    10 KB (1,861 words) - 09:47, 17 October 2021
  • ...<math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = ...coefficients yields <math>p = \tfrac32</math>, and comparing the constant coefficients yields <math>-7 = pq + pr - qr = \tfrac32(q+r) - qr</math>. The fact that <
    6 KB (1,015 words) - 20:16, 28 December 2023
  • ...th>y^4 + (s + u)y^3 + (t + v + su)y^2 + (sv + tu)y + tv</math>. [[Equating coefficients]] gives the following [[system of equations]]:
    13 KB (2,369 words) - 14:56, 23 October 2024
  • ...>2</math> and <math>m</math> into the expression for <math>p(x)</math> and equating them, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. Rearranging, we have < ...</math> which also equals to <math>4+2a+b</math> from the second equation. Equating both, we have <math>4a + b + 12 = 0.</math> We can easily count that there
    5 KB (954 words) - 16:14, 22 December 2024
  • Let <math>P(x)</math> be a polynomial with rational coefficients such that when <math>P(x)</math> is divided by the polynomial the squares of the coefficients of that polynomial?
    10 KB (1,749 words) - 15:05, 28 October 2024
  • A quick test reveals that <math>2</math> is a root of the equation. Comparing coefficients we can factorize the equation into: ...\cdot x^{-2}</math> is <math>f'(x)=1-27\cdot x^{-2}+46\cdot x^{-3}</math>. Equating <math>f'(x)</math> to zero, we get<cmath>1-27\cdot x^{-2}+46\cdot x^{-3}=0
    15 KB (2,475 words) - 22:18, 31 July 2024
  • Equating coefficients gives <math>A_1 = 2</math> and <math>A_1 + A_2 = 0</math>, so <math>A_2 = -
    4 KB (572 words) - 09:05, 15 July 2024