Question

Factorize each of the following polynomials:
 \[{x^3} + 13{x^2} + 31x - 45\] given that \[x + 9\] is a factor.

Answer 1

Hint: In this question, we have directly been given a factor of the polynomial. So, we just need to divide the given polynomial by the given factor. Then, when the division is completed, we are going to have a quotient which further is going to need factorization as on dividing a polynomial of degree 3 by another polynomial of degree 1, we get a polynomial of degree 2 (following the rule of division of number with same bases but different exponents) and a degree 2 polynomial still has two factors. And it is upon factoring that degree 2 polynomial, we are going to arrive at our final answer.
Formula Used:
For this question, we use the concept of division of a number with equal bases but unequal powers:
And it says:
 \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]

Complete step-by-step answer:
Here, we are given that the polynomial \[f\left( x \right) = {x^3} + 13{x^2} + 31x - 45\] has a factor \[x + 9\] .
So, we divide \[f\left( x \right) = {x^3} + 13{x^2} + 31x - 45\] by \[x + 9\] .
 \[x + 9\mathop{\left){\vphantom{1\begin{array}{l}{x^3} + 13{x^2} + 31x - 45\\\dfrac{{{x^3} + 9{x^2}}}{{0{x^3} + 4{x^2}}}\\{\rm{ 4}}{x^2} + 31x\\{\rm{ }}\dfrac{{{\rm{4}}{x^2} + 36x}}{{0{x^2} - 5x}}\\{\rm{ }} - 5x - 45\\{\rm{ }}\dfrac{{ - 5x - 45}}{0}\end{array}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^3} + 13{x^2} + 31x - 45\\\dfrac{{{x^3} + 9{x^2}}}{{0{x^3} + 4{x^2}}}\\{\rm{ 4}}{x^2} + 31x\\{\rm{ }}\dfrac{{{\rm{4}}{x^2} + 36x}}{{0{x^2} - 5x}}\\{\rm{ }} - 5x - 45\\{\rm{ }}\dfrac{{ - 5x - 45}}{0}\end{array}}}}
\limits^{\displaystyle\,\,\, {{x^2} + 4x - 5}}\]
So, the other factor of \[f\left( x \right)\] is \[{x^2} + 4x - 5\] .
Now, we will factorize \[{x^2} + 4x - 5\] by splitting the middle term, and we have:
 \[{x^2} + 4x - 5 = {x^2} + 5x - x - 5 = x\left( {x + 5} \right) - 1\left( {x + 5} \right) = \left( {x - 1} \right)\left( {x + 5} \right)\]
Hence, \[{x^3} + 13{x^2} + 31x - 45 = \left( {x - 1} \right)\left( {x + 5} \right)\left( {x + 9} \right)\]
So, the correct answer is “ $\left( {x - 1} \right)\left( {x + 5} \right)\left( {x + 9} \right)$ ”.

Note: So, we saw that in questions like these, when we are given a factor, we need to first divide the bigger polynomial by the given factor. Then, if the quotient obtained on division is not a polynomial of order 1, then we are going to have to still factorize the quotient polynomial. Then when the quotient’s polynomial is factorized to the point where each factor is of order 1, then only we are going to reach the final answer, which is the factorization of the original polynomial.