Question

How do you factor $5{{x}^{3}}-320$?

Answer 1

This type of problem is based on the concept of factoring a polynomial. First, we have to consider the polynomial with degree 3. First, we have to look for any common term which can be a variable or constant. Take 5 common from both terms of the polynomial. We know that 64 is the cube of 4. Using the identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$, we can further simplify the polynomial. Here a=x and b=4. Then, consider ${{x}^{2}}+4x+16$ and find the factors separately.

Complete step by step solution:
According to the question, we are asked to find the factors of $5{{x}^{3}}-320$.
We have been given the polynomial is $5{{x}^{3}}-320$. ---------(1)
The given polynomial is of degree 3 and variable x.
To find the factors, we have to find the common term which can be a variable or constant.
We can express the polynomial as
$5{{x}^{3}}-320=5{{x}^{3}}-5\times 64$
Here, we find that 5 are common in the two terms of the polynomial.
On taking 5 common, we get
$5{{x}^{3}}-320=5\left( {{x}^{3}}-64 \right)$
We know that 64 is the cube of 4.
Therefore, we get
$5{{x}^{3}}-320=5\left( {{x}^{3}}-{{4}^{3}} \right)$
We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.
Let us use the identity in the expression.
Here, a=x and b=4.
$5{{x}^{3}}-320=5\left( x-4 \right)\left( {{x}^{2}}+4x+{{4}^{2}} \right)$
We know that square of 4 is 16.
$\Rightarrow 5{{x}^{3}}-320=5\left( x-4 \right)\left( {{x}^{2}}+4x+16 \right)$ -------------(2)
Let us consider ${{x}^{2}}+4x+16$.
We know that for a quadratic polynomial $a{{x}^{2}}+bx+c$, the factors are $\left( x-\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right)$.
On comparing with the considered expression, we get
a=1, b=4 and c=16.
On substituting the values, we get
${{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 1\times 16}}{2\times 1} \right)$
On further simplification, we get
${{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 16}}{2} \right)$
$\Rightarrow {{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{16-4\times 16}}{2} \right)$
$\Rightarrow {{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{16-64}}{2} \right)$
$\Rightarrow {{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{-48}}{2} \right)$
We know that 48 can be written as the product of 16 and 3.
$\Rightarrow {{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{-16\times 3}}{2} \right)$
We know that $\sqrt{ab}=\sqrt{a}\sqrt{b}$. Using this property, we get
${{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm \sqrt{16}\sqrt{-3}}{2} \right)$
We know that $\sqrt{16}=4$. We get
${{x}^{2}}+4x+16=\left( x-\dfrac{-4\pm 4\sqrt{-3}}{2} \right)$
Let us take 2 common from the numerator.
$\Rightarrow {{x}^{2}}+4x+16=\left( x-\dfrac{2\left( -2\pm 2\sqrt{-3} \right)}{2} \right)$
We find that 2 are common in the numerator and denominator. On cancelling 2, we get
${{x}^{2}}+4x+16=\left( x-\left( -2\pm 2\sqrt{-3} \right) \right)$
We know that the negative term under root is an imaginary number.
Therefore, we get $\sqrt{-3}=\sqrt{3}i$, where I is a complex number.
Therefore, we get
${{x}^{2}}+4x+16=\left( x-\left( -2\pm 2\sqrt{3}i \right) \right)$
Therefore, we can write the expression as
${{x}^{2}}+4x+16=\left( x-\left( -2-2\sqrt{3}i \right) \right)\left( x-\left( -2+2\sqrt{3}i \right) \right)$
On removing the inner bracket, we get
${{x}^{2}}+4x+16=\left( x+2+2\sqrt{3}i \right)\left( x+2-2\sqrt{3}i \right)$ ------------(3)
On substituting expression (3) in expression (2), we get
$5{{x}^{3}}-320=5\left( x-4 \right)\left( x+2+2\sqrt{3}i \right)\left( x+2-2\sqrt{3}i \right)$
Therefore, the factors of $5{{x}^{3}}-320$ are 5(x-4), $\left( x+2+2\sqrt{3}i \right)$ and $\left( x+2-2\sqrt{3}i \right)$.

Note: Whenever we get such a type of problem, we should always compare the given polynomial with known identities and formulas. Simplify and then solve. Avoid calculation mistakes based on sign convention. Since the given polynomial is of degree 3, we get 3 factors.