Question

given that 3x³+2x²-3x+4 = (AX+B)(X-1)(X+2)+C(X-1)+D find the all value of A, B, C, and D​

Answer 1

Answer:

A=3, B=-1, C=4, D=6

Step-by-step explanation:

3x³+2x²-3x+4 = (Ax+B)(x-1)(x+2)+C(x-1)+D

= (Ax+B)(x²+x-2)+Cx-C+D

= Ax³+Ax²-2Ax+Bx²+Bx-2B+Cx-C+D

= Ax³+(A+B)x²+(-2A+B+C)x+(-2B-C+D)

By comparison,

A=3.

A+B=2, 3+B=2, B=-1.

-2A+B+C=-3, -6-1+C=-3, C=4.

-2B-C+D=4, 2-4+D=4, D=6.

Answer 2

Answer:

(ax^2-ax+bx-b) (x+2) +cx-c +d

ax^3-ax^2+bx^2-bx+2ax^2-2ax+2bx-2b+cx-c+d

comparing coefficients of x^3, x^2, x

we get

3=a

a+b=2

b=-1

b+c-2a=-3

-1+c-6=-3

c=4

-2b-c+d=4

2-4+d=4

d=6