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asked Apr 12, 2023 133k views

1 Answer

Sagar Chamling · Answer 1 · 2023-04-18T06:22:25+0000

Expand the right side.

A(x+2)(x-4) = A(x^2 - 2x - 8) = Ax^2 - 2Ax - 8A

B(x-4)(x-1) = B(x^2-5x+4) = Bx^2-5Bx+4B

C(x-1)(x+2) = C(x^2+x-2) = Cx^2 + Cx - 2C

Then

4x^2 + 4x - 26 = (A + B + C)x^2 + (-2A - 5B + C)x + (-8A + 4B - 2C)

Two polynomials are equal if the coefficients on like-power terms are equal, so

$\begin{cases} A + B + C = 4 \\ -2A - 5B + C = 4 \\ -8A + 4B - 2C = -26 \end{cases}$

Solve the system of equations.

Eliminating
, we have

$(A + B + C) - (-2A - 5B + C) = 4 - 4 \\\\ ~~~~ \implies 3A + 6B = 0 \\\\ ~~~~ \implies A + 2B = 0$

$2(A + B + C) + (-8A + 4B - 2C) = 8 - 26 \\\\ ~~~~ \implies -6A + 6B = -18 \\\\ ~~~~ \implies -A + B = -3$

Eliminating
,

$(A + 2B) - 2(-A + B) = 0 + 6 \\\\ ~~~~ \implies 3A = 6 \\\\ ~~~~ \implies \boxed{A = 2}$

Solve for
and
.

$A + 2B = 0 \\\\ ~~~~ \implies B = -\frac A2 \\\\ ~~~~ \implies \boxed{B = -1}$

$A + B + C = 4 \\\\ ~~~~ \implies C = 4 - A - B \\\\ ~~~~ \implies \boxed{C = 3}$