Question:
Find $\alpha, \beta$ if $(x+1)$ and $(x+2)$ are the factors of $x^{3}+3 x^{2}-2 a x+\beta$
Solution:
Given, $f(x)=x^{3}+3 x^{2}-2 \alpha x+\beta$ and the factors are $(x+1)$ and $(x+2)$
From factor theorem, if they are the factors of f(x) then results of f(-2) and f(-1) should be zero
Let, x + 1 = 0
⟹ x = -1
Substitute value of x in f(x)
$f(-1)=(-1)^{3}+3(-1)^{2}-2 \alpha(-1)+\beta$
= −1 + 3 + 2α + β
= 2α + β + 2 ... 1
Let, x + 2 = 0
⟹ x = -2
Substitute value of x in f(x)
$f(-2)=(-2)^{3}+3(-2)^{2}-2 \alpha(-2)+\beta$
= −8 + 12 + 4α + β
= 4α + β + 4 .... 2
Solving 1 and 2 i.e (1 - 2)
⟹ 2α + β + 2 - (4α + β + 4) = 0
⟹ −2α - 2 = 0
⟹ 2α = −2
⟹ α = −1
Substitute α= -1 in equation 1
⟹ 2(−1) + β = -2
⟹ β = -2 + 2
⟹ β = 0
The values are α = −1 and β = 0