Question

How do you factor ${{x}^{2}}-13x+42$? \[\]

Answer 1

Hint: We recall polynomial and its factorization. We recall the factorisation of quadratic polynomial $a{{x}^{2}}+bx+c$ using splitting the middle term method where we find two number $p,q$ such that $p+q=b$ and $pq=c\times a$. We prime factorization of $c\times a$ to get $p,q$ and then take common necessary terms to factorize.

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 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 know that the general form quadratic polynomial is given by $a{{x}^{2}}+bx+c$. We factorize $a{{x}^{2}}+bx+c$ by splitting the middle term $b=p+q$ where $p\times q=c\times a$. We find $p,q$ as the factors from the prime factorization of $c\times a$. We are asked to find the factors of
\[{{x}^{2}}-13x+42\]
We compare with a general quadratic polynomial in one variable $a{{x}^{2}}+bx+c$ to have $a=1,b=-13,c=42$. So we find $p,q$ such that $p+q=-13,pq=1\times 42=42$. We prime factorize 42 to have
\[42=2\times 3\times 7\]
We see that $42=6\times 7$. If we take $p=6,q=7$ we get $pq=42$ but not $p+q=-13$. So we take a native sign on $6,7$ and assign $p=-6,q=-7\Rightarrow pq=\left( -6 \right)\left( -7 \right)=42$ and $p+q=\left( -6 \right)+\left( -7 \right)=-13$. So we can write the given polynomial as
\[\begin{align}
  & {{x}^{2}}-13x+42 \\
 & \Rightarrow {{x}^{2}}-\left( 6+7 \right)x+42 \\
 & \Rightarrow {{x}^{2}}-6x-7x+42 \\
\end{align}\]
We take $x$ common from first two terms and 7 common from last two terms to have;
\[\Rightarrow x\left( x-6 \right)-7\left( x-6 \right)\]
We take $\left( x-6 \right)$ common from both terms to have;
\[\Rightarrow \left( x-6 \right)\left( x-7 \right)\]
The above form is required factored form where factors are $\left( x-6 \right),\left( x-7 \right)$.\[\]

Note:
We note that if ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{k}}\left( x \right)$ are $k$ factors of $p\left( x \right)$ then we say $p\left( x \right)={{p}_{1}}\left( x \right){{p}_{2}}\left( x \right)...{{p}_{k}}\left( x \right)$ is factored completely if none of the factors ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right),...,{{p}_{k}}\left( x \right)$ can be factored further. Here ${{x}^{2}}-13x+42=\left( x-6 \right)\left( x-7 \right)$ is a complete factorization. We can alternatively use $\left( x-a \right)\left( x-b \right)={{x}^{2}}-\left( a+b \right)x+ab$ to factorize directly.