Question

How do you factor $3{{x}^{4}}-81x{{y}^{3}}$ ?

Answer 1

Hint: In the given function, if we common out a factor from both terms, then we can get an equation of the form ${{a}^{3}}-{{b}^{3}}$ . To factorize ${{a}^{3}}-{{b}^{3}}$ , we use the postulate $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ . With the help of this after evaluating, we can find the other factor/s.

Complete step by step solution:
The given expression is $3{{x}^{4}}-81x{{y}^{3}}$
Now let us first take the common terms from both terms.
As we can look through, we see that the term $x$ is common in both terms.
After taking the common factor out we get,
$\Rightarrow x(3{{x}^{3}}-81{{y}^{3}})$
From the given equation, considering the coefficient of the second term, we can factorize it as follows.
$81=3\times 3\times 3\times 3={{3}^{4}}$
Now let us consider the coefficients in front of the variables.
All the coefficients have a common factor $3$ .
Let us take that also out.
After taking the common factor out we get,
$\Rightarrow 3x\left( {{x}^{3}}-{{3}^{3}}{{y}^{3}} \right)$
This is also the equation in the perfect cube form.
$\Rightarrow 3x\left( {{x}^{3}}-{{\left( 3y \right)}^{3}} \right)$
Now consider the second term.
We know that $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
Comparing the equation with the standard equation ${{a}^{3}}-{{b}^{3}}$ , we get
$a=x$ and $b=3y$
Substituting the values, we get,
$\Rightarrow \left( {{x}^{3}}-{{\left( 3y \right)}^{3}} \right)=\left( x-3y \right)\left( {{x}^{2}}+3xy+{{\left( 3y \right)}^{2}} \right)$
On further evaluation we get,
$\Rightarrow \left( x-3y \right)\left( {{x}^{2}}+3xy+9{{y}^{2}} \right)$
Now writing it all together we get,
$\Rightarrow 3x\left( x-3y \right)\left( {{x}^{2}}+3xy+9{{y}^{2}} \right)$
Hence the factors for the polynomial $3{{x}^{4}}-81x{{y}^{3}}$ are $3x\left( x-3y \right)\left( {{x}^{2}}+3xy+9{{y}^{2}} \right)$

Note: This postulate given here can be proved by the principle of mathematical induction. Here, we can also directly use the proven formula $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ to factorize the given equation, but mistakes might occur if you do not remember the equation perfectly. You can always cross-check your answer by placing the value of $x$ (factors of the expression) back in the equation. If you get LHS = RHS then your answer is correct.