Polynomial

Question

0

288

1

If (x + 2)(3x^2 - x + 8) = Ax^3 + Bx^2 + Cx + D, what is the value of A + 2B + 3C + 4D?

Guest May 4, 2022

+9675

+1

Best Answer

Expand out.

$\quad(x + 2)(3x^2 - x + 8)\\ = x(3x^2 - x + 8) + 2(3x^2 - x + 8)\\ = 3x^3 - x^2 + 8x + 6x^2 - 2x + 16\\ = 3x^3 + 5x^2 + 6x + 16$

Now you can compare the coefficients of $3x^3 + 5x^2 + 6x + 16$ and $Ax^3 + Bx^2 + Cx + D$ to find A, B, C, D. Then just calculate A + 2B + 3C + 4D.

MaxWong May 4, 2022

MaxWong · Answer 1 · 2022-05-04T05:11:52

Expand out.

$\quad(x + 2)(3x^2 - x + 8)\\ = x(3x^2 - x + 8) + 2(3x^2 - x + 8)\\ = 3x^3 - x^2 + 8x + 6x^2 - 2x + 16\\ = 3x^3 + 5x^2 + 6x + 16$

Now you can compare the coefficients of $3x^3 + 5x^2 + 6x + 16$ and $Ax^3 + Bx^2 + Cx + D$ to find A, B, C, D. Then just calculate A + 2B + 3C + 4D.