Question

The polynomial ax^3+3x^2-3 and 2x^2-5x+a when divided by (x-4) leave the remainder R1 and R2.find the value of a in this case. R1+R2=0

Answer 1

when $ax^2+3x^2-3$ divided by (x-4)
x-4=0
x=-4

$R_1 =a(-4)^3 + 3(-4)^2-3$
$R_1 =64a+48-3$
$R_1 =64a+45$

when $2x^2-5x+a$ divided by (x-4)
x-4=0
x=-4

$R_2 =2(-4)^2-5(-4)+a$
$R_2 =32+20+a$
$R_2 =52+a$
since $R_1 + R_2 = 0$

$64a+45+52+a=0$
$65a+97=0$

$a= \frac{-97}{65}$