Question

How do you factor ${{x}^{3}}+3{{x}^{2}}-9x-27$ ? \[\]

Answer 1

Hint: We recall Euclidean division of polynomials and factor of a polynomial. We take ${{x}^{2}}$ common from the first two terms of the polynomial and $-9$ common from the last two terms of the given polynomial. We take $x+3$ common to find the factorization. We use algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factorize completely. \[\]

Complete step by step answer:
We know that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$.We use Euclidean division formula and can write as
\[ p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right)\]
We also know that if the remainder polynomial is zero that is $r\left( x \right)=0$ then we call $d\left( x \right),q\left( x \right)$ factor polynomial of $p\left( x \right)$ or simply factors of $p\left( x \right)$. We denote the given polynomial $p\left( x \right)$ and have
\[p\left( x \right)={{x}^{3}}+3{{x}^{2}}-9x-27\]
We observe if we can take common. We see that we can take ${{x}^{2}}$ common from first two terms and $-9$ common from last two terms to have;
\[\Rightarrow p\left( x \right)={{x}^{2}}\left( x+3 \right)-9\left( x+3 \right)\]
We take $x+3$ common from both the terms in the above step to have;
\[\Rightarrow p\left( x \right)=\left( x+3 \right)\left( {{x}^{2}}-9 \right)\]
 We see that we have $p\left( x \right)$ factorized but not completely factored since $\left( {{x}^{2}}-9 \right)$ can be factored further. We use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=x,b=3$ to factorize ${{x}^{2}}-9={{x}^{2}}-{{3}^{2}}$ to have;
 \[\begin{align}
  & \Rightarrow p\left( x \right)=\left( x+3 \right)\left( x+3 \right)\left( x-3 \right) \\
 & \Rightarrow p\left( x \right)={{\left( x+3 \right)}^{2}}\left( x-3 \right) \\
\end{align}\]

The above expression is the required factorization of the given polynomial $p\left( x \right)={{x}^{3}}+3{{x}^{2}}-9x-27$.\[\]

Note: We note that $p\left( x \right)={{p}_{1}}\left( x \right){{p}_{2}}\left( x \right)...{{p}_{n}}\left( x \right)$ is a complete factorization if none of the factors ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...{{p}_{n}}\left( x \right)$ can be factored further. If $p\left( a \right)=0$ then $x=a$is zero of the polynomial. We can alternatively factorize if we can guess a zero and use factor theorem which states that if $p\left( a \right)=0$ then $x-a$ is a factor of $p\left( x \right)$. We then divide $p\left( x \right)$ by $x-a$ to get the other factors. Since the given polynomial is cubic (degree 3) accordingly we have obtained quadratic factor ${{\left( x+3 \right)}^{2}}$ and linear factor $x-3$.