Calculus - complex number?

Question

the quadratic x^2-4 is a factor of 3x^3+aX^2+bx-8. find the values of a and b.

Answer 1

Given equation : f(x) =  3x³ + ax² + bx - 8 = 0

The quadratic factor = two factors = x² - 4

x² - 4 = ( x - 2)(x + 2)

two factors = ( x - 2)(x + 2)

(x + 2) is factor of f(x).So substitute x = -2.

f(-2) =  3(-2)³ + a(-2)² + b(-2) - 8 = 0

-24 + 4a - 2b - 8 = 0

4a - 2b = 32

2a - b = 16

b =2a - 16

( x - 2) is factor of f(x).So substitute x = 2.

f(2) =  3(2³) + a(2²) + b2 - 8 = 0

24 + 4a + 2b - 8 = 0

4a + 2b = -16

2a + b = -8

Substitute b = 2a - 16

2a + 2a - 16 = -8

4a = 8

a = 2

b = 2a - 16 = 2(2) - 16 = 4 - 16

b = -12

Solution is a = 2 and b = -12