Question

Question 15
Check whether p(x) is a multiple of g(x) or not:

(i) $p\left(x\right)=x^3–5x^2+4x–3,g\left(x\right)=x–2$

(ii) $p\left(x\right)=2x^3–11x^2–4x+5,g\left(x\right)=2x+1$

Answer 1

$g\left(x\right)=x-2$ [given]
Then, zero of g(x) is 2.
p(2) is remainder, using Remainder Theorem.
Now,
$p\left(2\right)=(2{)}^3–5(2{)}^2+4\left(2\right)–3$ $[\because p(x)=x^3-5x^2+4x-3,given]$
$=8–20+8–3=-7\ne 0$
Since, remainder $\ne 0$, p(x) is not a multiple of g(x).

(ii) Here, $g\left(x\right)=2x+1$
Then, zero of g(x) is $\frac{-1}{2}$.
p$(\frac{-1}{2}$) is remainder, using Remainder Theorem.
Now,
$p\left(\frac{-1}{2}\right)=2{\left(\frac{-1}{2}\right)}^3-11{\left(\frac{-1}{2}\right)}^2-4\left(\frac{-1}{2}\right)+5$ $[\because p(x)=2x^3-11x^2-4x+5]$
$=2\left(\frac{-1}{8}\right)-11\left(\frac{1}{4}\right)+2+5=\frac{-1}{4}-\frac{11}{4}+7$
$=\frac{-1-11+28}{4}=\frac{16}{4}=4$

Since Remainder $\ne$ 0, p(x) is not a multiple of g(x).