Question

How do you factor $40{x^3} + 5$?

Answer 1

Hint:Factoring reduces the higher degree equation into its reduced equation. In the above given question, we need to reduce a third power polynomial with the help of the standard formula of sum of two cubes. We need to write the above two parts in the form of addition of two cubes.

Complete step by step solution:
The above equation is a third power polynomial or it is even called a cubic polynomial which includes at least one monomial that has the highest power of 3. To solve the equation, we will use the standard formula of difference of cubes because the equation contains negative signs.

\[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]

So, to bring the equation in this format we need to find out the terms a and b in the expression.
Since 40 and 5 are not perfect cubes we need to reduce the equation to get perfect cubes. So,5 can be divided by both terms. We can take 5 common from the equation.

\[5(8{x^3} + 1)\]

The term \[8{x^3}\] can be written as \[{(2x)^3}\] ,because the cube root of 8 is

Hence, 2x=a

The next term 1 can be written as \[{(1)^3}\],because the cube root of 1 is 1.

Therefore, 1=b

So, further by rewriting the equation in terms of \[{a^3} + {b^3}\] we get,
\[
5({(2x)^3} + {(1)^3}) \\
= 5(2x + 1)({(2x)^2} - (2x)(1) + (1)) \\
\\
\]

Further, applying the multiplicative distributive property on the expression which is:
\[{(xy)^a} = {x^a}{y^a}\]
Therefore,
\[
\Rightarrow 5(2x + 1)({2^2}{x^2} - 2x \times 1 + {1^2}) \\
\Rightarrow 5(2x + 1)(4{x^2} - 2x + 1) \\

\]

Hence, the final solution we get is the above equation.

Note: It is always important while solving a third power polynomial to understand that the equation belong to which different factoring scenarios like whether it is sum of two cubes or difference of two cubes and identifying a and b in such a way that the terms should be perfect cubes.