Question

(A) Given that the expression​​ x^3-ax^2+bx+c leaves the same ​remainder when divided by x+1 or x-2, find a in term of b.

Answer 1

Hello,

A:

$\begin{array}{c|ccc|c}&x^3&x^2&x&1\\&1&-a&b&c\\x=-1&&-1&a+1&-a-b-1\\---&---&---&---&---\\&1&-a-1&a+b+1&-a-b+c-1\\\end{array}\\\\\\\begin{array}{c|ccc|c}&x^3&x^2&x&1\\&1&-a&b&c\\x=2&&2&-2a+4&-4a+2b+8\\---&---&---&---&---\\&1&-a+2&-2a+b+4&-4a+2b+c+8\\\end{array}\\\\\\\\-a-b+c-1=-4a+2b+c+8\\\\\boxed{b=a-3}$

B:

$(2x-1)^3+6(3+4x^2)\\\\=8x^3-12x^2+6x-1+18+24x^2\\\\=8x^3+12x^2+6x+17\\\\\\\begin{array}{c|ccc|c}&x^3&x^2&x&1\\&8&12&6&17\\x=-\dfrac{1}{2} &&-4&-4&-1\\---&---&---&---&---\\&8&8&2&16\\\end{array}\\\\\\(2x-1)^3+6(3+4x^2)=(2x+1)(4x^2+4x+1)+16$