Question

Given that x-2 and x +1 are factors of $f\left(x\right)x^3+3x^2+ax+b;$ calculate the values of a and b. Hence, find all the factors of f(x).

Answer 1

Using factor theorem:
x+1 = 0 or x = -1
x+2 = 0 or x = -2
so, p(-1) = 0
or ${(-1)}^3$ + 3${(-1)}^2$ - 3a (-1) + b = 0
or -1 + 3 + 3a + b = 0
or 3a + b = -2 -----(1)

Also p(-2) = 0
or ${(-2)}^3$ + 3${(-2)}^2$ - 3a(-2) + b = 0
-8 + 12 + 6a + b = 0
or 6a + b = -4 --------(2)
(1) $\times$ 2 gives 6a + 2b = -4 ---------(3)
(3) - (1) gives
6a + 2b - (6a+b) = -4 - (-4)
6a + 2b -6a -b = -4 + 4
b = 0
Putting b = 0 in (2) gives
6a + 0 = -4
a = $\frac{-4}{6}$
a = $\frac{-2}{3}$