Question

if 2x^2+ax^2-bx-15 has 2x+3 as a factor and leaves a remainder-5 when divided by (x-1),find the values of a and b.

Answer 1

Let(2x+3)=0
2x=-3
x=-3/2
f(x)=2x^2+ax^2-bx-15
f(-3/2)=2*(-3/2)^2+a*(-3/2)^2-b*(-3/2)-15
=18/4+9a/4+3b/2-15
=(18+9a+6b-60)/4
=(-42+9a+6b)/4
given:(2x+3) is a factor. Therefore, remainder=0
So,(-42+9a+6b)/4=0
-42+9a+6b=0
9a+6b=42
3(3a+2b)=3(14)
3a+2b=14_eq.1
Now,let(x-1)=0
x=1
f(x)=2x^2+ax^2-bx^2-15
f(1)=2+a-b-15
=-13+a-b
Acc.to ques: remainder=-5
So,-13+a-b=-5
a-b=-5+13
a-b=8_eq.2
now multiply eq.2 by 2, to get:
2a-2b=16_eq.3
on adding eq.1 and eq.2, we get:5a=30
a=6 and b=(14-3*a)/2
b=(14-18)/2
b=-2