Question

How do you factor $f\left( x \right)={{x}^{4}}-12{{x}^{3}}+59{{x}^{2}}-138x+130$?

Answer 1

Hint: In this question we have been given with a polynomial equation which has a degree $4$. We cannot solve this expression by direct methods since it does not have any common term in all the present terms. we can see that the expression is similar to the expansion of the term ${{\left( x-3 \right)}^{4}}$. We will use the formula of expansion which is ${{\left( a-b \right)}^{4}}={{a}^{4}}-4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}-4a{{b}^{3}}+{{b}^{4}}$. We will then simplify the equation and factor it by taking terms common and then use the expansion formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$to get the final solution.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow f\left( x \right)={{x}^{4}}-12{{x}^{3}}+59{{x}^{2}}-138x+130$
Since there is no direct way to calculate the factors of the given equation, we will simplify the expression by writing it in terms of the expansion of the term ${{\left( x-3 \right)}^{4}}$
We know that ${{\left( a-b \right)}^{4}}={{a}^{4}}-4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}-4a{{b}^{3}}+{{b}^{4}}$ therefore, on using the expansion formula, we get:
${{\left( x-3 \right)}^{4}}={{x}^{4}}-12{{x}^{3}}+54{{x}^{2}}-108x+81$
Now on comparing both the expressions we can see that:
$\Rightarrow {{x}^{4}}-12{{x}^{3}}+59{{x}^{2}}-138x+130=\left( {{x}^{4}}-12{{x}^{3}}+54{{x}^{2}}-108x+81 \right)+\left( 5{{x}^{2}}-30x+49 \right)$
Therefore, the function can be written as:
$\Rightarrow {{\left( x-3 \right)}^{4}}+\left( 5{{x}^{2}}-30x+49 \right)$
We can split and write the expression as:
$\Rightarrow {{\left( x-3 \right)}^{4}}+\left( 5{{x}^{2}}-30x+45+4 \right)$
On taking $5$ common from the terms, we get:
$\Rightarrow {{\left( x-3 \right)}^{4}}+\left( 5\left( {{x}^{2}}-6x+9 \right)+4 \right)$
Now we will use the expansion of ${{\left( x-3 \right)}^{2}}$ which is ${{x}^{2}}-6x+9$. On substituting it in the above expression, we get:
$\Rightarrow {{\left( x-3 \right)}^{4}}+\left( 5{{\left( x-3 \right)}^{2}}+4 \right)$
On opening the brackets, we get:
$\Rightarrow {{\left( x-3 \right)}^{4}}+5{{\left( x-3 \right)}^{2}}+4$
Now consider ${{\left( x-3 \right)}^{2}}=a$ on substituting it in the above expression, we get:
$\Rightarrow {{a}^{2}}+5a+4$
On factorizing, we get:
$\Rightarrow \left( a+1 \right)\left( a+4 \right)$
On resubstituting the value of $a$, we get:
$\Rightarrow \left( {{\left( x-3 \right)}^{2}}+1 \right)\left( {{\left( x-3 \right)}^{2}}+4 \right)$
Now we can write $1=-{{i}^{2}}$ and $4={{\left( -2i \right)}^{2}}$, on substituting, we get:
$\Rightarrow \left( {{\left( x-3 \right)}^{2}}-{{i}^{2}} \right)\left( {{\left( x-3 \right)}^{2}}-{{\left( 2i \right)}^{2}} \right)$
On using the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, we get:
$\Rightarrow \left( x-3-i \right)\left( x-3+i \right)\left( x-3-2i \right)\left( x-3+2i \right)$, which is the required solution.

Note: In this polynomial expression, we do not have real factors which is why we could not use direct methods such as actual division to solve it. The term $i$ in the factors represents a complex number which is the square root of the negative one. It is written as $i=\sqrt{-1}$. The various expansion formulas should be remembered while doing these types of sums.