Question

How do you factor the given polynomial completely: ${{a}^{2}}-100$?

Answer 1

Hint: We start solving the problem by making the necessary arrangements in the given polynomial. We then make use of the result that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to proceed through the problem. We then make the necessary calculations to get the factorization result of the polynomial ${{a}^{2}}-100$.

Complete step by step answer:
According to the problem, we are asked to factorize the given polynomial ${{a}^{2}}-100$ completely.
We have given the polynomial ${{a}^{2}}-100$.
We can see that ${{a}^{2}}-100={{\left( a \right)}^{2}}-{{\left( 10 \right)}^{2}}$ ---(1).
From equation (1), we can see that the R.H.S (Right Hand Side) resembles the form ${{a}^{2}}-{{b}^{2}}$. We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. Let us use this result in equation (1).
$\Rightarrow {{a}^{2}}-100=\left( a-10 \right)\left( a+10 \right)$.
So, we have found the factors of the polynomial ${{a}^{2}}-100$ as $\left( a-10 \right)$ and $\left( a+10 \right)$.
$\therefore $ The factorization result of the polynomial ${{a}^{2}}-100$ are $\left( a-10 \right)\left( a+10 \right)$.

Note:
We can also solve this problem by first finding the zeros of the given polynomial which will give the required factors. We can also find the sum product of zeroes of the given polynomial.
We can also solve the given problem as shown below:
We have given the polynomial: ${{a}^{2}}-100$.
$\Rightarrow {{a}^{2}}-100={{a}^{2}}-10a+10a-100$.
$\Rightarrow {{a}^{2}}-100=\left( a\times a \right)+\left( a\times -10 \right)+\left( 10\times a \right)+\left( 10\times -10 \right)$ ---(2).
Let us now take the common factors out by grouping the first two terms and next two terms from equation (2).
\[\Rightarrow {{a}^{2}}-100=a\times \left( a-10 \right)+10\times \left( a-10 \right)\] ---(3).
From equation (3), we can see that both the terms have a common factor $\left( a-10 \right)$. So, let us take it common.
\[\Rightarrow {{a}^{2}}-100=\left( a+10 \right)\left( a-10 \right)\].
Similarly, we can expect problems to factorize the polynomial $36{{x}^{2}}-1$ completely.