Question

Express the factors of $\left( {{x}^{2}}-3x+5 \right)\left( {{x}^{2}}+3x+5 \right)$ as a difference of two squares.

Answer 1

Hint: We find the zeroes(say $\alpha ,\beta $) of the both the quadratic polynomials $\left( {{x}^{2}}-3x+5 \right)$ and $\left( {{x}^{2}}+3x+5 \right)$ using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and express each of them in the factorized from $\left( x-\alpha \right)\left( x-\beta \right)$. We use the algebraic identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ as difference of two squares. \[\]

Complete step by step answer:
We know that if a quadratic equation in general form $a{{x}^{2}}+bx+c=0,a\ne 0$ has roots $\alpha ,\beta $ then we can express the quadratic polynomial in $a{{x}^{2}}+bx+c=0$ as
\[a{{x}^{2}}+bx+c=\left( x-\alpha \right)\left( x-\beta \right)\]
The roots of the quadratic equation are obtained by factorization when the discriminant $D={{b}^{2}}-4ac>0$. If the discriminant $D={{b}^{2}}-4ac<0$ then we get complex roots of the quadratic equation which we can find directly using the formula
\[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a},\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
Let us denote the given expression of a polynomial in the question as $p\left( x \right)$ . So we have
\[p\left( x \right)=\left( {{x}^{2}}-3x+5 \right)\left( {{x}^{2}}+3x+5 \right)\]
We see that the given expression $p\left( x \right)$ has two factors $\left( {{x}^{2}}-3x+5 \right)$ and $\left( {{x}^{2}}+3x+5 \right)$. We denote ${{p}_{1}}\left( x \right)={{x}^{2}}-3x+5$ and ${{p}_{2}}\left( x \right)={{x}^{2}}+3x+5$.

The discriminant for first quadratic factor ${{p}_{1}}\left( x \right)={{x}^{2}}-3x+5=0$ is $D={{\left( -3 \right)}^{2}}-4\cdot 1\cdot 5=-11<0$. So we find the roots of ${{p}_{1}}\left( x \right)=0$ using the quadratic formula directly and get the roots ${{\alpha }_{1}},{{\beta }_{1}}$ as,
\[{{\alpha }_{1}}=\dfrac{3+\sqrt{-11}}{2},{{\beta }_{1}}=\dfrac{3-\sqrt{-11}}{2}\]
So we can express ${{p}_{1}}\left( x \right)$ as
\[{{p}_{1}}\left( x \right)=\left( x-{{\alpha }_{1}} \right)\left( x-{{\beta }_{1}} \right)=\left( x-\dfrac{3+\sqrt{-11}}{2} \right)\left( x-\dfrac{3-\sqrt{-11}}{2} \right)...(1)\]
The discriminant for the second quadratic factor ${{p}_{2}}\left( x \right)={{x}^{2}}+3x+5=0$ is $D={{\left( 3 \right)}^{2}}-4\cdot 1\cdot 5=-11<0$. So we find the roots of ${{p}_{2}}\left( x \right)=0$ using the quadratic formula directly and get the roots ${{\alpha }_{2}},{{\beta }_{2}}$ as,
\[{{\alpha }_{2}}=\dfrac{-3+\sqrt{-11}}{2},{{\beta }_{2}}=\dfrac{-3-\sqrt{-11}}{2}\]
So we can express ${{p}_{2}}\left( x \right)$ as
\[\begin{align}
  & {{p}_{1}}\left( x \right)=\left( x-{{\alpha }_{2}} \right)\left( x-{{\beta }_{2}} \right)=\left( x-\dfrac{\left( -3+\sqrt{-11} \right)}{2} \right)\left( x-\dfrac{\left( -3-\sqrt{-11} \right)}{2} \right) \\
 & \Rightarrow {{p}_{2}}\left( x \right)=\left( x+\dfrac{3-\sqrt{-11}}{2} \right)\left( x+\dfrac{3+\sqrt{-11}}{2} \right)...(2) \\
\end{align}\]
We replace the obtained values for ${{p}_{1}}\left( x \right)$ from equation (1) and ${{p}_{2}}\left( x \right)$ from equation (2) in $p\left( x \right)$. So we have
\[\begin{align}
  & p\left( x \right)={{p}_{1}}\left( x \right){{p}_{2}}\left( x \right) \\
 & \Rightarrow p\left( x \right)=\left( x-\dfrac{3+\sqrt{-11}}{2} \right)\left( x-\dfrac{3-\sqrt{-11}}{2} \right)\left( x+\dfrac{3-\sqrt{-11}}{2} \right)\left( x+\dfrac{3+\sqrt{-11}}{2} \right) \\
 & \Rightarrow p\left( x \right)=\left( x-\dfrac{3+\sqrt{-11}}{2} \right)\left( x+\dfrac{3+\sqrt{-11}}{2} \right)\left( x+\dfrac{3-\sqrt{-11}}{2} \right)\left( x-\dfrac{3-\sqrt{-11}}{2} \right) \\
\end{align}\]
Let us use the algebraic identity of $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ and proceed
\[\begin{align}
  & \Rightarrow p\left( x \right)=\left( {{x}^{2}}-{{\left( \dfrac{3+\sqrt{-11}}{2} \right)}^{2}} \right)\left( {{x}^{2}}-{{\left( \dfrac{3-\sqrt{-11}}{2} \right)}^{2}} \right) \\
 & \Rightarrow p\left( x \right)=\left( {{\left( x \right)}^{2}}-{{\left( \dfrac{3+i\sqrt{11}}{2} \right)}^{2}} \right)\left( {{\left( x \right)}^{2}}-{{\left( \dfrac{3-i\sqrt{11}}{2} \right)}^{2}} \right) \\
\end{align}\]
The above result is expressed as difference of two squares where $i=\sqrt{-1}$ is an imaginary number.\[\]

Note:
We note that the roots equal when $D=0$ . The roots are distinct when . $D>0,D\ne 0$The roots are rational when $D$ is a perfect square. The roots are integral and distinct when $D>0,D\ne 0$, $D$ is a perfect square and $2a$ divides $D$ exactly. The pair of values $\dfrac{3+11i}{2},\dfrac{3-11i}{2}$ are called conjugates.