Question

How do you factor completely $ {{x}^{3}}-9x $ ?

Answer 1

Hint: We first keep the variables in one side to factorise. The first process in factorisation is to take common terms out. We take $ x $ as the common term. Then we form the equation according to the identity $ {{a}^{2}}-{{b}^{2}} $ to form the factorisation of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . We place values $ a=x;b=3 $ .

Complete step-by-step answer:
We need to find the factorised form of $ {{x}^{3}}-9x $ . It’s a cubic equation of $ x $ .
First, we take $ x $ as the common term and get $ {{x}^{3}}-9x=x\left( {{x}^{2}}-9 \right) $ .
Now we take the quadratic equation $ {{x}^{2}}-9 $ . We need to factorise it.
The simplified form of
 $ {{x}^{2}}-9 $ is $ {{x}^{2}}-9={{x}^{2}}-{{3}^{2}} $ .
Now we find the factorisation of the equation $ {{x}^{2}}-{{3}^{2}}=0 $ using the identity of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . We place the values $ a=x;b=3 $ .
Therefore, we get
$ {{x}^{2}}-{{3}^{2}}=\left( x+3 \right)\left( x-3 \right) $ .
The final factorised form of
$ {{x}^{3}}-9x $ is $ {{x}^{3}}-9x=x\left( x+3 \right)\left( x-3 \right) $ .
In case we are solving the given polynomial for roots, we can also apply the quadratic equation formula to solve the equation $ {{x}^{2}}-9 $ .
We know for a general equation of quadratic $ a{{x}^{2}}+bx+c=0 $ , the value of the roots of x will be
$ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
The solution for $ {{x}^{3}}-9x=0 $ will be $ x=0,3,-3 $ as we get
$ {{x}^{3}}-9x=x\left( x+3 \right)\left( x-3 \right)=0 $ .
Therefore, $ x\left( x+3 \right)\left( x-3 \right)=0 $ has multiplication of three polynomials giving a value of 0. This means at least one of them has to be 0.
So, values of x are $ x=0,3,-3 $ .
So, the correct answer is “ $ x\left( x+3 \right)\left( x-3 \right)=0 $”.

Note: The highest power of the variable or the degree of a polynomial is always equal to the summation of the highest degrees of those factorised polynomials.
In case of $ {{x}^{3}}-9x $ the polynomial was cubic. The summation of the highest degrees of those factorised polynomials $ {{x}^{3}}-9x=x\left( x+3 \right)\left( x-3 \right) $ is also $ 1+1+1=3 $ .