Question

How do you factor ${{x}^{4}}+27x$?

Answer 1

Hint: Factoring is a process of finding the factors of the given expression with which it is formed. For example, the factors of the arithmetic value 21 are values 3 and 7. These factors when multiplied together will give the value 21. Similarly, we can find the factors of an algebraic expression.

Complete Step by Step Solution:
The given algebraic expression is ${{x}^{4}}+27x$. This is a polynomial expression of the fourth degree having a single variable x. The term polynomial indicates that it has variables, coefficients, and constants.
By analyzing the equation, we can see that there is an exponential $\left( {{x}^{4}} \right)$. One common approach is that we can try to reduce the value of the exponent in the expression and then group them together into smaller factors.
Let us take out the common factor in the expression.
 $\Rightarrow {{x}^{4}}+27x=x\left( {{x}^{3}}+27 \right)$ ……(1)
At this point, we don’t have enough components to attempt the grouping process. We should add some terms so that we can find some common factors to take out. We know that
$\Rightarrow {{\left( a+b \right)}^{3}}=\left( a+b \right){{\left( a+b \right)}^{2}}=\left( a+b \right)\left( {{a}^{2}}+2ab+{{b}^{2}} \right)$
$\Rightarrow {{\left( a+b \right)}^{3}}={{a}^{3}}+2{{a}^{2}}b+a{{b}^{2}}+b{{a}^{2}}+2a{{b}^{2}}+{{b}^{3}}$
After simplifying, we get,
$\Rightarrow {{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$
If we assume that $a=x$ and $b=3$, the expression can be rewritten as
$\Rightarrow {{\left( x+3 \right)}^{3}}={{x}^{3}}+9{{x}^{2}}+27x+27$
If we compare the above equation with the expression $\left( {{x^3} + 27} \right)$ in equation (1), we can see that the missing terms are $9{x^2}$ and $\;27$.
Therefore the equation (1) can be written as
$\Rightarrow {x^4} + 27x = x\left( {{x^3} + 3{x^2} - 3{x^2} - 9x + 9x + 27} \right)$
The terms $\left( {3{x^2} - 3{x^2}} \right)$ and $\left( { - 9x + 9x} \right)$ are equal to zero and so the final result will not be affected.
Now let us attempt to group the components
$\Rightarrow {{x}^{4}}+27x=x\left( {{x}^{2}}\left( x+3 \right)-3x\left( x+3 \right)+9\left( x+3 \right) \right)$
In the above equation, we can take out $(x+3)$
$\Rightarrow {{x}^{4}}+27x=x\left( \left( x+3 \right){{x}^{2}}-3x+9 \right)$
$\Rightarrow {{x}^{4}}+27x=x\left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)$

The expression $x\left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)$ is the factored form of ${{x}^{4}}+27x$.

Note:
The quadratic expression $\left( {{x}^{2}}-3x+9 \right)$ cannot be factored further. If we factorize the constant, we get $3\times 3$. We cannot write $-3x$ in terms of the factors of the constant 9. And so, we will be unable to group this algebraic expression. Hence $x\left( x+3 \right)\left( {{x}^{2}}-3x+9 \right)$ is the final factored form of ${{x}^{4}}+27x$.