Question

How do you factor the trinomial ${{x}^{2}}-6x-16$ ?

Answer 1

Hint: First try to find the two numbers whose summation is $-6$ and multiplication is $-16$. Then do the necessary factorization as per the procedure. Split $-6x$ as $2x-8x$ and take $\left( x+2 \right)$ common from both the parts. On the very next step $\left( x-8 \right)$ common can be taken out.

Complete step by step answer:
As we know for the factorization of the polynomial of form $a{{x}^{2}}+bx+c$ we need to find two numbers whose summation is ‘b’ and multiplication is ‘c’.
Now for our trinomial ${{x}^{2}}-6x-16$, we need to find two numbers whose summation is $-6$ and multiplication is $-16$.
So the numbers are 2 and $-8$
Whose summation$=2+\left( -8 \right)=2-8=-6$
And multiplication $=2\times \left( -8 \right)=-16$
Hence the trinomial ${{x}^{2}}-6x-16$ can be factorized as
$\begin{align}
  & {{x}^{2}}-6x-16 \\
 & \Rightarrow {{x}^{2}}+2x-8x-16 \\
 & \Rightarrow x\left( x+2 \right)-8\left( x+2 \right) \\
 & \Rightarrow \left( x+2 \right)\left( x-8 \right) \\
\end{align}$
This is the required solution of the given question.

Note: This is not the only method for solving such questions. These types of questions can also be solved by completing the square method in which we need to convert the whole polynomial into some square factors as required.
$\begin{align}
  & {{x}^{2}}-6x-16 \\
 & \Rightarrow {{\left( x \right)}^{2}}-2\cdot x\cdot 3+{{\left( 3 \right)}^{2}}-{{\left( 3 \right)}^{2}}-16 \\
\end{align}$
As we know ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ so, ${{\left( x \right)}^{2}}-2\cdot x\cdot 3+{{\left( 3 \right)}^{2}}$can be written as ${{\left( x-3 \right)}^{2}}$
$\begin{align}
  & \Rightarrow {{\left( x-3 \right)}^{2}}-9-16 \\
 & \Rightarrow {{\left( x-3 \right)}^{2}}-25 \\
\end{align}$
25 can be written as ${{\left( 5 \right)}^{2}}$
$\Rightarrow {{\left( x-3 \right)}^{2}}-{{\left( 5 \right)}^{2}}$
Again as we know ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ so, our expression can be written as
$\begin{align}
  & \Rightarrow \left( x-3+5 \right)\left( x-3-5 \right) \\
 & \Rightarrow \left( x+2 \right)\left( x-8 \right) \\
\end{align}$
This is the alternative method. In this method also we are getting the same answer. But since this is a lengthy method so it will take more time and calculation to get the required solution. Hence the above method is not preferable over the method mentioned earlier.