Question

Find the value of a and b so that
$8 {x}^{4} + 14{x}^{3} - 2 {x}^{2} + ax + b$
is exactly divisible by
$4 {x}^{2} + 3x - 2$
​

Answer 1

8x^4+14x^3–2x^2+ax+b

=2x^2(4x^2+3x-2)+8x^3+2x^2+ax+b

=2x^2(4x^2+3x-2)+2x(4x^2+3x-2)-4x^2+4x+ax+b

=2x^2(4x^2+3x-2)+2x(4x^2+3x-2)-[4x^2-(4+a)x-b]

f(x) is exactly divisible by g(x) , therefore

g(x)=4x^2-(4+a)x-b

4x^2+3x-2=4x^2-(4+a)x-b

or 3x-2= -(4+a)x-b , equating both sides the

coeff. of x and constant term.

3=-(4+a)

3=-4-a

a=-4–3

a= -7 ,Answer

-2=-b

b=2 , Answer