Question

Show that:
$2x-3$ Is a factor of $x+2{{x}^{3}}-9{{x}^{2}}+12$.

Answer 1

Hint: This type of question is based on factor theorem. To check whether $2x-3$ is a factor of$x+2{{x}^{3}}-9{{x}^{2}}+12$ we will use factor theorem which state that a polynomial $f\left( x \right)$ has a factor $\left( x-a \right)$ if and only if $f\left( a \right)=0$. So firstly we will find the value of $a$ by using the factor given and then we will substitute it in the equation. Then we will simplify the expression to check whether it is satisfied or not.

Complete step by step solution:
The equation given to us is as follows:
$x+2{{x}^{3}}-9{{x}^{2}}+12$.
Firstly we will let both the equation and the factor equal to some function as:
$p\left( x \right)=x+2{{x}^{3}}-9{{x}^{2}}+12$…..$\left( 1 \right)$
$g\left( x \right)=2x-3$……$\left( 2 \right)$
Now we will put the equation (2) equal to zero and get,
$g\left( x \right)=2x-3=0$
$\begin{align}
  & \Rightarrow 2x-3=0 \\
 & \Rightarrow 2x=3 \\
 & \Rightarrow x=\dfrac{3}{2} \\
\end{align}$
For $g\left( x \right)$ to be a factor of $p\left( x \right)$ $x=\dfrac{3}{2}$ should satisfy the equation below:
$p\left( \dfrac{3}{2} \right)=0$
So equating $x=\dfrac{3}{2}$ in equation (1) and simplifying we get,
$\begin{align}
  & \Rightarrow p\left( \dfrac{3}{2} \right)=\dfrac{3}{2}+2\times {{\left( \dfrac{3}{2} \right)}^{3}}-9{{\left( \dfrac{3}{2} \right)}^{2}}+12 \\
 & \Rightarrow p\left( \dfrac{3}{2} \right)=\dfrac{3}{2}+2\times \dfrac{27}{8}-9\times \dfrac{9}{4}+12 \\
 & \Rightarrow p\left( \dfrac{3}{2} \right)=\dfrac{3}{2}+\dfrac{27}{4}-\dfrac{81}{4}+12 \\
\end{align}$
Taking L.C.M on right side and simplifying we get,
$\begin{align}
  & \Rightarrow p\left( \dfrac{3}{2} \right)=\dfrac{6+27-81+48}{4} \\
 & \Rightarrow p\left( \dfrac{3}{2} \right)=\dfrac{0}{4} \\
 & \therefore p\left( \dfrac{3}{2} \right)=0 \\
\end{align}$
So we get $p\left( \dfrac{3}{2} \right)=0$
So$g\left( x \right)$ is a factor of $p\left( x \right)$

Hence $2x-3$ is a factor of $x+2{{x}^{3}}-9{{x}^{2}}+12$.

Note: Factor theorem states that if a polynomial f(x) of degree $n\ge 1$ and ‘a’ is any real number then, if f (a) = 0 then only (x-a) is a factor of the polynomial. The two problems where Factor Theorem is usually applied are when we have to factorize a polynomial and also when we have to find the roots of the polynomial. It is also used to remove known zeros from a polynomial while leaving all unknown zeros intact. It is a special case of a polynomial remainder theorem.