Question

If the polynomial 5x cube +4x square+31x+a is exactly divisible by x-2 ,then find the value of a ,hence factories the polynomial

Answer 1

Answer:

$\bf{5x^3+4x^2+31x-118=(x-2)(x^2+14x+59)}$

Step-by-step explanation:

$\text{Let P(x)}=5x^3+4x^2+31x+a$

$\text{since P(x) is exactly divisible by (x-2), by factor theorem P(2)=0}$

$5(2)^3+4(2)^2+31(2)+a=0$

$\implies\:40+16+62+a=0$

$\implies\:118+a=0$

$\implies\:a=-118$

$\therefore\:P(x)=5x^3+4x^2+31x-118$

Now,

$5x^3+4x^2+31x-118=(x-2)(x^2+px+59)$

Equating coefficient of x on both sides , we get

31=-2p+59

31-59=-2p

-28=-2p

p=14

$\implies\:\bf{5x^3+4x^2+31x-118=(x-2)(x^2+14x+59)}$