Question

How do you factor \[{x^4} + 6{x^3} - 36{x^2} - 216x\]

Answer 1

Hint: Here we will try to simply the above polynomial equation by finding the factor of the equation which we will find from the root value of the polynomial, and then solve a simplified quadratic equation to get the final answer, which is not in the form of the power of $x$.

Complete step-by-step solution:
The given polynomial is:
\[ \Rightarrow {x^4} + 6{x^3} - 36{x^2} - 216x\]
Since $x$ is common in all the terms we will simplify the equation by taking it out as common, on taking $x$ out as common we get:
\[ \Rightarrow x({x^3} + 6{x^2} - 36x - 216)\]
Now we will try to deduce the factor of the given equation.
First, when $x = 0$ the equation becomes:
\[ \Rightarrow 6({0^3} + 6{(0)^2} - 36(0) - 216)\]
Now on simplifying we get:
\[ \Rightarrow 6( - 216)\]
On multiply we get
\[ \Rightarrow - 1296\] Which is not equal to $0$ therefore, $x = 0$ is not a root of the equation
Now when $x = 1$ the equation becomes:
\[ \Rightarrow 6({1^3} + 6{(1)^2} - 36(1) - 216)\]
Now on simplifying we get:
\[ \Rightarrow 6(1 + 6 - 36 - 216)\]
Let us add and subtract the term and multiply we get
\[ \Rightarrow - 1470\] which is not equal to $0$ therefore, $x = 1$ is not a root of the equation.
Now when $x = 2$ the equation becomes:
\[ \Rightarrow 6({2^3} + 6{(2)^2} - 36(2) - 216)\]
Now on simplifying we get:
\[ \Rightarrow 6(8 + 24 - 72 - 216)\]
Let us add and subtract the term and multiply we get
\[ \Rightarrow - 1536\] Which is not equal to $0$ therefore, $x = 2$ is not a root of the equation
Now when $x = 3$ the equation becomes:
\[ \Rightarrow 6({3^3} + 6{(3)^2} - 36(3) - 216)\]
Now on squaring the term we get:
\[ \Rightarrow 6(27 + 54 - 108 - 216)\]
On simplifying we get
\[ \Rightarrow - 1458\] which is not equal to $0$ therefore, $x = 3$ is not a root of the equation.
Now when $x = 4$ the equation becomes:
\[ \Rightarrow 6({4^3} + 6{(4)^2} - 36(4) - 216)\]
Now on squaring the term we get:
\[ \Rightarrow 6(64 + 64 - 144 - 216)\]
On simplifying we get
\[ \Rightarrow - 1312\] which is not equal to $0$ therefore, $x = 4$ is not a root of the equation.
Now when $x = 5$ the equation becomes:
\[ \Rightarrow 6({5^3} + 6{(5)^2} - 36(5) - 216)\]
Now on squaring and multiply the term we get:
\[ \Rightarrow 6(125 + 150 - 180 - 216)\]
On simplify we get
\[ \Rightarrow - 726\] which is not equal to $0$ therefore, $x = 5$ is not a root of the equation.
Note that when $x = 6$ the equation becomes:
\[ \Rightarrow 6({6^3} + 6{(6)^2} - 36(6) - 216)\]
Now on simplifying we get:
\[ \Rightarrow 6(216 + 216 - 216 - 216)\]
Which is:
\[ \Rightarrow 6(0)\] which is equal to $0$.
Since $x = 6$ is a root of the polynomial equation, $x - 6 = 0$ is a factor therefore the given equation can be written as:
\[ \Rightarrow x({x^2}(x - 6) + 12x(x - 6) + 36(x - 6))\]
On taking the term $(x - 6)$ as common, we get:
$ \Rightarrow x(x - 6)({x^2} + 12x + 36)$
Now since the latter term is in the form of a quadratic equation, we will solve it by splitting up the middle term, the equation can be written as:
$ \Rightarrow x(x - 6)({x^2} + 6x + 6x + 36)$
On taking the common terms we get:
$ \Rightarrow x(x - 6)(x(x + 6) + 6(x + 6))$
Now since the term $(x + 6)$ is common in both the terms we can take it out as common:
$ \Rightarrow x(x - 6)((x + 6)(x + 6)$
On simplifying we get:
$ \Rightarrow x(x - 6){(x + 6)^2}$

$ x(x - 6){(x + 6)^2}$ is the required answer.

Note: It is to be remembered that a polynomial equation is a combination of variables and their coefficients, the equation given above is a $4th$ degree polynomial equation.
The factors of a polynomial equation are the terms of degree $1$, which have to be multiplied among themselves so that we get the required polynomial equation. In the question above, $x,(x - 6)$ and $(x + 6)$ are factors of the given polynomial.