Question

If 3x3+2x2-3x+4=(Ax+B)X-1)x+2)+Cx-1)+D for all values of x,then A+B+C+D is​

Answer 1

Answer:

The value of A+B+C+D is 12

Step-by-step explanation:

$3x^3+2x^2-3x+4=(Ax+B)(x-1)(x+2)+C(x-1)+D$......(1)

put x=1in (1)

$3(1)^3+2(1)^2-3(1)+4=(A(1)+B)(1-1)(1+2)+C(1-1)+D$

$2+4=D$

$D=6$

put x=-2 in (1)

$3(-2)^3+2(-2)^2-3(-2)+4=(A(-2)+B)(x-1)(-2+2)+C(-2-1)+D$

$-24+8+6+4=C(-2-1)+D$

$-6=C(-2-1)+6$

$-6-6=C(-3)$

$-12=C(-3)$

$C=4$

put x=0 in (1)

$3(0)^3+2(0)^2-3(0)+4=(A(0)+B)((0)-1)(0+2)+4(0-1)+6$

$4=-2B-4+6$

$4-2=-2B$

$2=-2B$

$B=-1$

put x=2 in (2)

$3(2)^3+2(2)^2-3(2)+4=(A(2)+B)(2-1)(2+2)+C(2-1)+D$

$24+8-6+4=(A(2)+B)(2-1)(2+2)+C(2-1)+D$

$30=4(2A+B)+C+D$

$30=8A+4B+C+D$

$30=8A+4(-1)+4+6$

$30=8A-4+10$

$24=8A$

$A=3$

Now,

$A+B+C+D$

$=3+(-1)+4+6$

$=12$