Question

How do you factor and solve ${{x}^{2}}-16=6x$?

Answer 1

Hint: The equation given in the above question has a degree equal to two, and therefore it is a quadratic equation. To solve it, we first have to write it in the standard form by making its RHS equal to zero. For this we need to subtract $6x$ from the given equation ${{x}^{2}}-16=6x$ to obtain the quadratic equation ${{x}^{2}}-6x-16=0$. Then, since we have to solve the given equation by factoring it, we will use the middle term splitting method. For this, we will split the middle term as $-6x=2x-8x$ to get the equation ${{x}^{2}}+2x-8x-16=0$ and then group the first two and the last two terms on the LHS. Then taking the common factors outside till the polynomial in the LHS is completely factored, the equation will be factored. Finally, using the zero product rule the equation will be solved.

Complete step by step solution:
The equation given in the above question is
\[\Rightarrow {{x}^{2}}-16=6x\]
The above quadratic equation is not written in the standard form. For writing it in the standard form, we subtract $6x$ from both the sides of the above equation to get
\[\begin{align}
  & \Rightarrow {{x}^{2}}-16-6x=6x-6x \\
 & \Rightarrow {{x}^{2}}-6x-16=0 \\
\end{align}\]
Now, for factoring, we use the middle term splitting method to split the middle term as $-6x=2x-8x$ in the above equation to get
$\Rightarrow {{x}^{2}}+2x-8x-16=0$
Now, we combine the first two and the last two terms as
$\Rightarrow \left( {{x}^{2}}+2x \right)+\left( -8x-16 \right)=0$
Now, we can take $x$ and $-8$ outside the two pairs to get
$\Rightarrow x\left( x+2 \right)-8\left( x+2 \right)=0$
Now, taking $\left( x+2 \right)$ outside, we get
$\Rightarrow \left( x+2 \right)\left( x-8 \right)=0$
Now, using the zero product rule, we get
$\begin{align}
  & \Rightarrow x+2=0 \\
 & \Rightarrow x=-2 \\
\end{align}$
And
$\begin{align}
  & \Rightarrow x-8=0 \\
 & \Rightarrow x=8 \\
\end{align}$
Hence, we have factored and solved the given equation and obtained the solutions as $x=-2$ and $x=8$

Note:
The middle term cannot be split in any arbitrary way. It can only be split into two terms such that the product of the two terms is equal to the product of the first and the third term. For example, in the case of the above solution, we split the middle term as $-6x=2x-8x$ so that the product of $2x$ and $-8x$, which is equal to $-16{{x}^{2}}$, is equal to the product of the first and the third terms.