How do you factor completely: $16{{x}^{2}}-9$?

Hint: To solve this question, the given polynomial $16{{x}^{2}}-9$ has to be written in the form of the difference of two squares so that we can apply the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor it. For writing it in the form of the difference of two squares, we have to put $16={{4}^{2}}$ and $9={{3}^{2}}$ in the given polynomial $16{{x}^{2}}-9$ to get ${{4}^{2}}{{x}^{2}}-{{3}^{2}}$. Then using the property of exponent given by ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, we can write ${{4}^{2}}{{x}^{2}}-{{3}^{2}}$ as ${{\left( 4x \right)}^{2}}-{{3}^{2}}$. Then on applying the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ the given polynomial will be completely factored.

Complete step by step solution:
Let us write the given polynomial as
$\Rightarrow p\left( x \right)=16{{x}^{2}}-9$
We know that $16={{4}^{2}}$ and $9={{3}^{2}}$. Putting these above, we get
\[\Rightarrow p\left( x \right)={{4}^{2}}{{x}^{2}}-{{3}^{2}}\]
Now, using the exponent property ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, we can write the above polynomial as
\[\Rightarrow p\left( x \right)={{\left( 4x \right)}^{2}}-{{3}^{2}}\]
Applying the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can factorise the above polynomial as
$\Rightarrow p\left( x \right)=\left( 4x+3 \right)\left( 4x-3 \right)$
Hence, the given polynomial is completely factored as $\left( 4x+3 \right)\left( 4x-3 \right)$.

Note:
Do not try to factorize the polynomial further by putting $4x={{\left( \sqrt{2x} \right)}^{2}}$ in the factor $\left( 4x-3 \right)$. This is because for making this substitution $x$ must be greater than or equal to zero, since the square root function is defined for non negative values only. But we do not know anything about the values of $x$. We can also factorize the given quadratic polynomial by finding its zeroes. For this we need to equate the given quadratic polynomial to zero to get the quadratic equation $16{{x}^{2}}-9=0$ and using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we can determine the zeros of the polynomial. Then by using the factor theorem, we will be able to factorize the polynomial.