Question

How do you factor $49{{x}^{2}}+112x+64?$

Answer 1

Hint: We divide the second term of the given polynomial into two equal terms. Then we group the terms accordingly. Later, we factor out the greatest common divisor.

Complete step by step solution:
We are given with the following polynomial to be factored,
$\Rightarrow 49{{x}^{2}}+112x+64.$
The second term, that is, the term with $x$, is $112x.$
Let us divide this term with $2.$
Then we will get, $\dfrac{112x}{2}=56x.$
That implies, $112x=56x+56x.$
So, from what we have obtained we can understand that,
$\Rightarrow 49{{x}^{2}}+112x+64=49{{x}^{2}}+56x+56x+64.$
Now, we can group the terms of the above obtained polynomial as,
$\Rightarrow 49{{x}^{2}}+112x+64=\left( 49{{x}^{2}}+56x \right)+\left( 56x+64 \right).$
Let us factor out the greatest common divisors from each group. In the first group, the greatest common divisor is $7x$ and in the second group, the greatest common divisor is $8.$
So, we will get the polynomial as,
$\Rightarrow 49{{x}^{2}}+112x+64=7x\left( 7x+8 \right)+8\left( 7x+8 \right).$
And there is another greatest common divisor in the polynomial we have obtained after factoring out $7x$ and $8.$
And that divisor is $7x+8.$
So, in the step, we are going to take this divisor out as,
$\Rightarrow 49{{x}^{2}}+112x+64=\left( 7x+8 \right)\left( 7x+8 \right).$
Now we are directed to,
$\Rightarrow 49{{x}^{2}}+112x+64={{\left( 7x+8 \right)}^{2}}.$
Hence, the factor is $7x+8.$.

Note: Note that we have done a division on $112x$ with $2.$ It is done based on the identity that is derived below:
Suppose that we have two values $a$ and \[b.\]
Let us, now, take the square of sum of these two variables.
That is, ${{\left( a+b \right)}^{2}}.$
Now, we can expand this square to get the following,
$\Rightarrow \left( a+b \right)\left( a+b \right).$
Let us multiply each term in the first parenthesis with each term in the second parenthesis.
We get,
$\Rightarrow {{a}^{2}}+ab+ab+{{b}^{2}}.$
Add the same term to get,
$\Rightarrow {{a}^{2}}+2ab+{{b}^{2}}.$
So, we get this identity, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}.$
Since $112x=2\times 56x,$ using the above identity we can write,
$\Rightarrow 49{{x}^{2}}+112x+6=49{{x}^{2}}+2\times 56x+64.$
We know that $49{{x}^{2}}={{\left( 7x \right)}^{2}}$ and $64={{8}^{2}},$
We get,
$\Rightarrow 49{{x}^{2}}+112x+64={{\left( 7x \right)}^{2}}+2\times 56x+{{8}^{2}}.$
And so,
$\Rightarrow 49{{x}^{2}}+112x+64={{\left( 7x+8 \right)}^{2}}.$