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

Question:

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

(i) p(x) = x3 – 5x2 + 4x – 3,  g(x) = x – 2.

(ii) p(x) = 2x3 – 11x2 – 4x+ 5,  g(x) = 2x + l

Thinking Process

(i) Firstly, find the zero of g(x) and then put the value of in p(x) and simplify it.

(ii) lf p(a) = Q then p(x) is a multiple of g(x) and f(p(a) # Q then p(x) is not a multiple of g(x) where ‘a’ is a zero of g (x)).

Solution:

(i) $\because$                $g(x)=x-2$                   [given]

Then, zero of $g(x)$ is 2 .

Now,                     $p(2)=(2)^{3}-5(2)^{2}+4(2)-3$              $\left[\because p(x)=x^{3}-5 x^{2}+4 x-3\right.$, given]

$=8-20+8-3=-7 \neq 0$

Since, remainder $\neq 0$, so $p(x)$ is not a multiple of $g(x)$.

(ii) Here, $g(x)=2 x+1$.

Then, zero of $g(x)$ is $\frac{-1}{2}$.

Now, $p\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)^{3}-11\left(\frac{-1}{2}\right)^{2}-4\left(\frac{-1}{2}\right)+5$             $\left[\because p(x)=2 x^{3}-11 / x^{2}-4 x+5\right]$

$=2\left(\frac{-1}{8}\right)-11\left(\frac{1}{4}\right)+2+5=\frac{-1}{4}-\frac{11}{4}+7$

$=\frac{-1-11+28}{4}=\frac{16}{4}=4$

Since, remainder $\neq 0$, so $p(x)$ is not a multiple of $g(x)$.