Question

Check whether $7+3x$ is a factor of $3x^3+7x$.

Answer 1

Step $1:$ Finding the zero of $7+3x$:

Let $g\left(x\right)=7+3x$ and $p\left(x\right)=3x^3+7x$

The zero of $g\left(x\right)=7+3x$ can be obtained by solving, $g\left(x\right)=0$

$\begin{array}{rcl}g\left(x\right)& =& 0\\ & \Rightarrow & 7+3x=0\\ & \Rightarrow & 3x=-7\\ & \Rightarrow & x=\frac{-7}{3}\end{array}$

The zero of $g\left(x\right)=7+3x$ is $x=\frac{-7}{3}$.

Step $2:$ Finding whether $7+3x$ is a factor of $3x^3+7x$ or not:

Now, $g\left(x\right)=7+3x$, will be a factor of $p\left(x\right)=3x^3+7x$, only, if $g\left(x\right)$ divides $p\left(x\right)$ leaving remainder zero, it means $p\left(\frac{-7}{3}\right)=0$

Therefore, by using remainder theorem, we get,

$\begin{array}{rcl}p\left(x\right)& =& 3x^3+7x\\ p\left(\frac{-7}{3}\right)& =& 3\times {\left(\frac{-7}{3}\right)}^3+7\times \left(\frac{-7}{3}\right)\\ & =& 3\times \left(\frac{-343}{27}\right)+\left(\frac{-49}{3}\right)\\ & =& \left(\frac{-343}{9}\right)+\left(\frac{-49}{3}\right)\\ & =& \frac{-343+(-49\times 3)}{9}\\ & =& \frac{-343-147}{9}\\ & =& \frac{-490}{9}\ne 0\end{array}$

Hence, $7+3x$ is not a factor of $3x^3+7x$