Question

Check whether$\mathrm{p}\left(\mathrm{x}\right)$ is a multiple of $\mathrm{g}\left(\mathrm{x}\right)$or not: $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^3-5{\mathrm{x}}^2+4\mathrm{x}-3,\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}-2$

Answer 1

Step 1: Find zero of g(x)

$\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^3-5{\mathrm{x}}^2+4\mathrm{x}-3$

$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}-2$

According to the given details

$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}–2$,

Then, zero of $\mathrm{g}\left(\mathrm{x}\right)$,

$\begin{array}{rcl}\mathrm{g}\left(\mathrm{x}\right)& =& 0\\ \mathrm{x}–2& =& 0\\ \mathrm{x}& =& 2\end{array}$

$\Rightarrow$Zero of $\mathrm{g}\left(\mathrm{x}\right)=2$

Step 2: Substitute the value of $\mathrm{x}$in $\mathrm{p}\left(\mathrm{x}\right)$

$\begin{array}{rcl}\mathrm{p}\left(2\right)& =& {\left(2\right)}^3–5{\left(2\right)}^2+4\left(2\right)–3\\ & =& 8–20+8–3\\ \because \mathrm{p}\left(2\right)& =& –7\ne 0\end{array}$

Hence,$\mathrm{p}\left(\mathrm{x}\right)$ is not the multiple of $\mathrm{g}\left(\mathrm{x}\right)$since the remainder $\ne 0$.