Question

What is the factored form of $2{{x}^{3}}+4{{x}^{2}}-x$ ?

Answer 1

Hint: To find the factored form of $2{{x}^{3}}+4{{x}^{2}}-x$ , we have to first equate this polynomial to f(x). Then, we have to take the common factor x outside. This will lead to a function $\Rightarrow f\left( x \right)=x\left( 2{{x}^{2}}+4x-1 \right)$ . Then, we have to factorize $2{{x}^{2}}+4x-1$ using quadratic formula and substitute these factors in f(x).

Complete step-by-step solution:
We have to find the factored form of $2{{x}^{3}}+4{{x}^{2}}-x$ . Let us equate the given polynomial to f(x).
$\Rightarrow f\left( x \right)=2{{x}^{3}}+4{{x}^{2}}-x$
We can see that x is a common term. So let us take x outside. Hence, we can write the above function as
$\Rightarrow f\left( x \right)=x\left( 2{{x}^{2}}+4x-1 \right)...\left( i \right)$
Now, we have to find the factors of $2{{x}^{2}}+4x-1$ . Let us use a quadratic formula. We know that for a polynomial in the form$a{{x}^{2}}+bx+c$ , the roots are given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here, we can see that $a=2,b=4,c=-1$ .
$\Rightarrow x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 2\times -1}}{2\times 2}$
Let us simplify the terms inside the root.
$\begin{align}
  & \Rightarrow x=\dfrac{-4\pm \sqrt{16+8}}{4} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{24}}{4} \\
\end{align}$
We know that $\sqrt{24}=2\sqrt{6}$ . Hence, the above result becomes
$\Rightarrow x=\dfrac{-4\pm 2\sqrt{6}}{4}$
Let us take the common factor 2 outside.
$\Rightarrow x=\dfrac{2\left( -2\pm \sqrt{6} \right)}{4}$
Let us cancel the common factor 2 from the numerator and the denominator.
$\Rightarrow x=\dfrac{-2\pm \sqrt{6}}{2}$
Let us split the signs.
$\Rightarrow x=\dfrac{-2+\sqrt{6}}{2},x=\dfrac{-2-\sqrt{6}}{2}$
Let us take the terms RHS to the LHS.
$\begin{align}
  & \Rightarrow x=\dfrac{-2}{2}+\dfrac{\sqrt{6}}{2},x=\dfrac{-2}{2}-\dfrac{\sqrt{6}}{2} \\
 & \Rightarrow x+\dfrac{2}{2}-\dfrac{\sqrt{6}}{2}=0,x+\dfrac{2}{2}+\dfrac{\sqrt{6}}{2}=0 \\
 & \Rightarrow x+\dfrac{2-\sqrt{6}}{2}=0,x+\dfrac{2+\sqrt{6}}{2}=0 \\
\end{align}$
Hence, the factors of $2{{x}^{2}}+4x-1$ are $\left( x+\dfrac{2-\sqrt{6}}{2} \right)\left( x+\dfrac{2+\sqrt{6}}{2} \right)$ .
Now, we can write (i) as
$\Rightarrow f\left( x \right)=x\left( x+\dfrac{2-\sqrt{6}}{2} \right)\left( x+\dfrac{2+\sqrt{6}}{2} \right)$
Hence, the factored form of $2{{x}^{3}}+4{{x}^{2}}-x$ is $x\left( x+\dfrac{2-\sqrt{6}}{2} \right)\left( x+\dfrac{2+\sqrt{6}}{2} \right)$ .

Note: Students must know how to factorize the polynomials. The commonly used methods to factorize are grouping, splitting the middle terms and quadratic formula. After finding the values of x using quadratic formula, students must find the factors using it by rearranging the terms.