Question

Factorize the following polynomial: ${{x}^{2}}-\dfrac{9}{16}$ \[\]

Answer 1

Hint: We recall the definition of polynomial and factoring polynomials. We factorize the given polynomial ${{x}^{2}}-\dfrac{9}{16}$ into two linear polynomials by using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=x,b=\dfrac{3}{4}$. \[\]

Complete step-by-step answer:
We know that a polynomial expression with real number ${{a}_{o}},{{a}_{1}},...,{{a}_{n}}$ and variable $x$ named as $p\left( x \right)$ can be written as
\[p\left( x \right)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+....+{{a}_{1}}x+{{a}_{0}}\]
We also know that the highest power on a variable is called degree which here is $n$. If ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{m}}\left( x \right)$ are the polynomial factors of $p\left( x \right)$ which means if we divide $p\left( x \right)$ by any one of the polynomials${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{m}}\left( x \right)$ then we will get remainder zero. We can factorize $p\left( x \right)$ in the following way;
\[p\left( x \right)={{p}_{1}}\left( x \right){{p}_{2}}\left( x \right)...{{p}_{m}}\left( x \right)\]
We note that if ${{n}_{1}},{{n}_{2}},...,{{n}_{m}}$are the degrees of ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{m}}\left( x \right)$ then we shall have;
\[{{n}_{1}}+{{n}_{2}}+...+{{n}_{m}}=n\]
We are given in the question to factorize the following polynomial: ${{x}^{2}}-\dfrac{9}{16}$. We see that the given polynomial is of degree since the highest power on variable $x$ is 2 . We also see the second term in the polynomial is fraction $\dfrac{9}{16}$whose numerator 9 is square of 3 and denominator 4 is a square 4. So we can write using the definition of exponents
\[\dfrac{9}{16}=\dfrac{{{3}^{2}}}{{{4}^{2}}}={{\left( \dfrac{3}{4} \right)}^{2}}\]

So we have
\[\Rightarrow {{x}^{2}}-\dfrac{9}{16}={{\left( x \right)}^{2}}-{{\left( \dfrac{3}{4} \right)}^{2}}\]
We see that the above expression is in the form of ${{a}^{2}}-{{b}^{2}}$ and we know the algebra that we can factorize ${{a}^{2}}-{{b}^{2}}$ as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. So let use this identity for $a=x$ and $b=\dfrac{3}{4}$ and have;
\[\Rightarrow {{x}^{2}}-\dfrac{9}{16}=\left( x+\dfrac{3}{4} \right)\left( x-\dfrac{3}{4} \right)\]
The above expression is in the required factored form.\[\]

Note: We see that the degree of the obtained two factors $x+\dfrac{3}{4},x-\dfrac{3}{4}$ is 1 and 1 respectively whose sum is $1+1=2$ which the degree is of ${{x}^{2}}-\dfrac{9}{16}$, the product polynomial. The polynomial of degree 2 is called quadratic polynomial. The values of $x$ for which $p\left( x \right)=0$ are called zeroes of the polynomial. We can alternatively solve using factor theorem by first finding a zero say $x=a$by trial and error to get one factor $\left( x-a \right)$then dividing $p\left( x \right)$ by $\left( x-a \right)$ by other factor.