Question

How do you factor cubic trinomials ${{x}^{3}}-7x-6$ ?

Answer 1

Hint: To factorize the polynomial ${{x}^{3}}-7x-6$ , we will consider the polynomial to be and equate the polynomial to 0. We will then use trial and error methods, where we will have to substitute random values for x so that the polynomial value equals to 0. From this method, we will obtain one factor. Then we will divide the given polynomial with this factor which will give a quotient that is a quadratic polynomial. Finally, we have to factorize this quadratic polynomial by splitting the middle terms.

Complete step-by-step solution:
We need to factorize ${{x}^{3}}-7x-6$ . The given question is a cubic polynomial. To factorize a cubic polynomial, first we have to consider the polynomial to be $p\left( x \right)$ and equate the polynomial to 0. Since the polynomial is cubic, we know that there will be 3 factors.
$\Rightarrow p\left( x \right)={{x}^{3}}-7x-6=0$
Now, let us use trial and error methods. In this method, we will substitute random values for x so that the polynomial value equals to 0.
Let us consider $x=-1$ .
$\Rightarrow p\left( 1 \right)={{\left( -1 \right)}^{3}}-7\times \left( -1 \right)-6=-1+7-6=0$
Hence, $x=-1$ is a root of the given polynomial.
Therefore, by factor theorem, $\left( x+1 \right)$ is a factor of ${{x}^{3}}-7x-6$ .
Now, let us divide ${{x}^{3}}-7x-6$ by $\left( x+1 \right)$ .
\[\begin{align}
  & x+1\overset{{{x}^{2}}-x-6}{\overline{\left){{{x}^{3}}+0{{x}^{2}}-7x-6}\right.}} \\
 & \text{ }{{x}^{3}}+{{x}^{2}} \\
 & \text{ }-\text{ }- \\
  & \text{ }\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \text{ }-{{x}^{2}}-7x-6 \\
 & \text{ }-{{x}^{2}}-x \\
 & \text{ }+\text{ +} \\
 & \text{ }\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \text{ }-6x-6 \\
 & \text{ }-6x-6 \\
 & \text{ }+\text{ }+ \\
 & \text{ }\_\_\_\_\_\_\_\_\_ \\
 & \text{ }0 \\
\end{align}\]
Hence, we can write the polynomial as $p\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-x-6 \right)$
Now, let us factorize the quadratic polynomial $\left( {{x}^{2}}-x-6 \right)$ by splitting the middle terms.
$\Rightarrow {{x}^{2}}-x-6={{x}^{2}}+2x-3x-6$
We splitted $-x$ into $2x-3x$ as $-x$ is the sum of these two terms. Also, the product of 2 and 3 is 6.
Now, let us take the common terms outside.
$\begin{align}
  & \Rightarrow x\left( x+2 \right)-3\left( x+2 \right) \\
 & =\left( x-3 \right)\left( x+2 \right) \\
\end{align}$
Hence, the factors of ${{x}^{3}}-7x-6$ are $\left( x-3 \right)\left( x+1 \right)\left( x+2 \right)$

Note: Students must keep in mind that number of factors of a polynomial will be equal to the degree of the polynomial. We can also solve this problem in an alternate method as described below:
$p\left( x \right)={{x}^{3}}-7x-6$
Let us add and subtract ${{x}^{2}}$ .
$\Rightarrow p\left( x \right)={{x}^{3}}+{{x}^{2}}-{{x}^{2}}-7x-6$
We can write $-7x$ as a sum of $-6x$ and $x$ .
$\Rightarrow p\left( x \right)={{x}^{3}}+{{x}^{2}}-{{x}^{2}}+x-6x-6$
Now, let us take the common terms outside in the following manner.
$\begin{align}
  & \Rightarrow p\left( x \right)={{x}^{2}}\left( x+1 \right)-x\left( x+1 \right)-6\left( x+1 \right) \\
 & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-x-6 \right)\left( x+1 \right) \\
\end{align}$
Remaining steps are similar to the solution.