Question

Question 2
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
$\left(i\right)p\left(x\right)=2x^3+x^2-2x-1,g\left(x\right)=x+1$
$\left(ii\right)p\left(x\right)=x^3+3x^2+3x+1,g\left(x\right)=x+2$
$\left(iii\right)p\left(x\right)=x^3-4x^2+x+6,g\left(x\right)=x-3$

Answer 1

(i) g(x) = x + 1, x = - 1 to be substituted in
$p\left(x\right)=2x^3+x^2-2x-1\Rightarrow p(-1)=2(-1{)}^3+(-1{)}^2-2(-1)-1=-2+1+2-1=0$
So, g(x) is a factor of p(x).

(ii) g(x) = x + 2, substitute x = - 2 in p(x).
$p\left(x\right)=x^3+3x^2+3x+1\Rightarrow p(-2)=(-2{)}^3+3(-2{)}^2+3(-2)+1=-8+12-6+1=-1$
So, g(x) is not a factor of p(x) as p(-2)$\ne$0

(iii) g(x) = x - 3, substitute x = 3 in p(x).
$p\left(x\right)=x^3-4x^2+x+6\Rightarrow p\left(3\right)=(3{)}^3-4(3{)}^2+3+6=27-36+3+6=0$
Therefore, g(x) is a factor of $x^3-4x^2+x+6$.