Question

How do you factor $4{{x}^{2}}+16x+16$?

Answer 1

Hint: We use a grouping method to find the factor of the problem. We take common terms out to form the multiplied forms. The middle term process breaks the polynomial into multiples of two smaller polynomials.

Complete step-by-step answer:
We apply the middle-term factoring or grouping to factorise the polynomial.
Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.
In the case of $4{{x}^{2}}+16x+16$, we break the middle term $16x$ into two parts of $8x$ and $8x$.
So, $4{{x}^{2}}+16x+16=4{{x}^{2}}+8x+8x+16$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases gives $64{{x}^{2}}$. The grouping will be done for $4{{x}^{2}}+8x$ and $8x+16$. We try to take the common numbers out.
For $4{{x}^{2}}+8x$, we take $4x$ and get $4x\left( x+2 \right)$. For $8x+16$, we take 8 and get $8\left( x+2 \right)$.
The equation becomes $4{{x}^{2}}+16x+16=4{{x}^{2}}+8x+8x+16=4x\left( x+2 \right)+8\left( x+2 \right)$.
Both the terms have $\left( x+2 \right)$ in common. We take that term again and get
$\begin{align}
  & 4{{x}^{2}}+16x+16 \\
 & =4x\left( x+2 \right)+8\left( x+2 \right) \\
 & =\left( x+2 \right)\left( 4x+8 \right) \\
 & =4\left( x+2 \right)\left( x+2 \right) \\
 & =4{{\left( x+2 \right)}^{2}} \\
\end{align}$
Therefore, the factorisation of $4{{x}^{2}}+16x+16$ is $4\left( x+2 \right)\left( x+2 \right)$.

Note: We can form the square form for ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$. We first take 4 common from $4{{x}^{2}}+16x+16$ and get $4{{x}^{2}}+16x+16=4\left( {{x}^{2}}+4x+4 \right)$. Then we take the assumptions for $a=x;b=2$ and get $\left( {{x}^{2}}+4x+4 \right)={{\left( x+2 \right)}^{2}}$.
Therefore, $4{{x}^{2}}+16x+16=4{{\left( x+2 \right)}^{2}}=4\left( x+2 \right)\left( x+2 \right)$.