Question

Show that:
(i) x + 3 is a factor of $69+11x–x^2+x^3$.

(ii) 2x – 3 is a factor of $x+2x^3–9x^2+12$.

Answer 1

(i) Let $p\left(x\right)=x^3–x^2+11x+69$
We have to show that, x + 3 is a factor of p(x).
i.e. p(-3) = 0 [Using Remainder Theorem]
Now, $p(-3)=(-3{)}^3–(-3{)}^2+11(-3)+69$
$=-27–9–33+69=-69+69=0$.
Therefore, x +3 is a factor.

(ii) $p\left(x\right)=2x^3-9x^2+x+12$
We have to show that, 2x - 3 is a factor of p(x).
i.e. $p\left(\frac{3}{2}\right)=0$ [Using Remainder Theorem]
$p\left(\frac{3}{2}\right)=2(\frac{3}{2}{)}^3-9(\frac{3}{2}{)}^2+\frac{3}{2}+12$
$=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+12$
$=\frac{81-81}{4}=0$
Hence, (2x-3) is a factor of p(x).