Question

Find the values of a and b so that the polynomial $(x^4+a{x}^3-7x^2-8x+b)$ is exactly divisible by (x + 2) as well as (x +3).

Answer 1

ANSWER:
Let:
$f\left(x\right)=x^4+a{x}^3-7x^2-8x+b$
Now,
x+2=0

⇒x=-2
By the factor theorem, we can say:
f(x) will be exactly divisible by (x+2) if f(-2)=0
Thus, we have:
$f(-2)=(-2{)}^4+a\times (-2{)}^3-7\times -(2{)}^2-8\times -2+b=16-8a-28+16+b=4-8a+b$
∴ f(-2)=0

⇒8a-b=4 ...(1)
Also,
x+3=0

⇒x=-3
By the factor theorem, we can say:
f(x) will be exactly divisible by x+3 if f(-3)=0.
Thus, we have:
$f(-3)=(-3{)}^4+a\times (-3{)}^3-7\times (-3{)}^2-8\times -3+b=81-27a-63+24+b=42-27a+b$
∴ f(-3)=0

⇒27a-b=42 ...(2)
Subtracting 1 from 2, we get:

⇒19a=38

⇒a=2
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12