Given that (x^2−4) is a factor of the polynomial f(x), where f(x) =4x^4+7x^3 +ax^2 +bx +8. Find the values of a and b and hence factorize f(x)completley. Find the set of values for which f(x)=0
+130116 If x^2 - 4 is a factor.....then (x - 2) and (x + 2) are factors
Which means that 2 and - 2 are roots
So
4(2)^4+7(2)^3 +a(2)^2 +b(2) +8 = 0
4(-2)^4+7(-2)^3 +a(-2)^2 + b(-2) +8 = 0
Add these and we have that
128 + 8a + 16 = 0 ⇒ 8a = -144 ⇒ a = - 18
Subtract them and we have that
112 + 4b = 0
28 + b = 0 ⇒ b = -28
So....the polynomial is
4x^2 + 7x^3 - 18x^2 - 28x + 8 and we can write this as
[4x^2 + 7x^3 - 2x^2] - 16x^2 - 28x + 8
x^2 (4x^2 + 7x - 2) - 4 ( 4x^2 + 7x - 2 )
(x^2 - 4) (4x^2 + 7x - 2)
(x^2 - 4) (4x - 1) (x + 2)
(x - 2) (x + 2) (4x - 1) (x + 2) = 0
The values that make this = 0 are x= 2, x = 1/4 and x = -2
