Question

How do you factor ${x^3} + 6{x^2} - x - 30$?

Answer 1

Hint: We will first find one root of the given function ${x^3} + 6{x^2} - x - 30$ using hit and trial method. Then, we will make one factor using it and divide the given equation by it and get the required factors.

Complete step by step solution:
We are given that we are required to factor ${x^3} + 6{x^2} - x - 30$.
Let us say $f(x) = {x^3} + 6{x^2} - x - 30$.
Putting $x = 2$ in the given equation $f(x) = {x^3} + 6{x^2} - x - 30$.
We will then obtain: $f(2) = {2^3} + 6 \times {2^2} - 2 - 30$
Simplifying the calculations, we will then obtain the following:-
$ \Rightarrow f(2) = 8 + 24 - 2 - 30 = 0$
Therefore, $x = 2$ which means $\left( {x - 2} \right)$ is a factor of $f(x)$.
Dividing the given equation $f(x) = {x^3} + 6{x^2} - x - 30$ by $\left( {x - 2} \right)$, we will then obtain the following equation:-
$ \Rightarrow x - 2)\overline {{x^3} + 6{x^2} - x - 30} $
Multiplying the divisor by ${x^2}$, we will then obtain the following equation with us:-
                  ${x^2}$
$ \Rightarrow x - 2)\overline {{x^3} + 6{x^2} - x - 30} $
                 ${x^3} - 2{x^2}$
                 $\overline {8{x^2} - x - 30} $
Multiplying the divisor by 8x now, we will then obtain the following equation:-
                  ${x^2} + 8x$
$ \Rightarrow x - 2)\overline {{x^3} + 6{x^2} - x - 30} $
                 ${x^3} - 2{x^2}$
                 $\overline {8{x^2} - x - 30} $
                 $8{x^2} - 16x$
                 $\overline {15x - 30} $
Multiplying the divisor by 15 now, we will then obtain the following equation:-
                  ${x^2} + 8x + 15$
$ \Rightarrow x - 2)\overline {{x^3} + 6{x^2} - x - 30} $
                 ${x^3} - 2{x^2}$
                 $\overline {8{x^2} - x - 30} $
                 $8{x^2} - 16x$
                 $\overline {15x - 30} $
                  $15x - 30$
                  _______
                        0
Thus, we can write the given polynomial ${x^3} + 6{x^2} - x - 30$ as follows:-
$ \Rightarrow {x^3} + 6{x^2} - x - 30 = \left( {x - 2} \right)\left( {{x^2} + 8x + 15} \right)$
We can write the latter equation as follows:-
$ \Rightarrow {x^3} + 6{x^2} - x - 30 = \left( {x - 2} \right)\left( {{x^2} + 3x + 5x + 15} \right)$
Taking x common from the first two terms in the latter factor, we will then obtain the following equation with us:-
$ \Rightarrow {x^3} + 6{x^2} - x - 30 = \left( {x - 2} \right)\left\{ {x\left( {x + 3} \right) + 5x + 15} \right\}$
Taking 5 common from the last two terms in the latter factor, we will then obtain the following equation with us:-
$ \Rightarrow {x^3} + 6{x^2} - x - 30 = \left( {x - 2} \right)\left\{ {x\left( {x + 3} \right) + 5\left( {x + 3} \right)} \right\}$
Taking (x + 3) common from the last two terms in the latter factor, we will then obtain the following equation with us:-

$ \Rightarrow {x^3} + 6{x^2} - x - 30 = \left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 5} \right)$

Thus, we have the required factors.

Note:
The students must notice that we have an alternate way of factoring the quadratic equation involved in it as well. The alternate way is as follows:-
The given equation is ${x^2} + 8x + 15$.
Using the quadratic formula given by if the equation is given by $a{x^2} + bx + c = 0$, its roots are given by the following equation:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Thus, we have the roots of ${x^2} + 8x + 15$ given by:
$ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt {64 - 60} }}{2}$
Hence, the roots are – 5 and – 3.
Thus the factors are (x + 3) and (x + 5).