Question

How do you factor ${x^3} - 16x$?

Answer 1

Hint:
The given expression ${x^3} - 16x$ consists of the variable ‘x’, which can be treated as any number and can be operated like the same. Take common $'x'$ and use the algebraic identity of the difference of the squares, i.e. ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ . For getting the required expansion put $a = x$ and $b = 4$ , then solve it further.

Complete step by step answer:
Here in this problem, we are given an expression ${x^3} - 16x$, which is in the form of the difference of the cube of one number and a product of constant and ‘x’. And using the algebraic properties we need to factor this expression.
In the expression ${x^3} - 16x$, ‘x’ here is a variable that does not have a constant value and $16$ which is a real number with a constant value. In mathematics, a variable is a symbol that functions as a placeholder for varying expressions or quantities and is often used to represent an arbitrary element of a set.
The factor of the given expression can be done using an algebraic identity. The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials.
Now, in the given expression, let’s take $x$ common
$ \Rightarrow {x^3} - 16x = x\left( {{x^2} - 16} \right)$
As we know that square of $4$ is ${4^2} = 16$. Let’s use this in the above polynomial
\[ \Rightarrow x\left( {{x^2} - 16} \right) = x\left( {{x^2} - {4^2}} \right)\] ….(A)
We know the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ , if we put $x = a$ and $b = 4$ then we will get:
$ \Rightarrow {x^2} - {4^2} = \left( {x - 4} \right)\left( {x + 4} \right)$
Now we can combine this result with the expression obtained in (A). This will give us:
\[ \Rightarrow x\left( {{x^2} - {4^2}} \right) = x\left( {x - 4} \right)\left( {x + 4} \right)\]
Therefore, the factor of the expression ${x^3} - 16x$ is given by ${x^3} - 16x = x\left( {x - 4} \right)\left( {x + 4} \right)$

Hence, we can say $x$ , $\left( {x - 4} \right)$ and $\left( {x + 4} \right)$ are the factors of ${x^3} - 16x$

Additional Information:
To understand the concept of expansion better, we can use another example where we can factor the expression ${x^3} + 11x$
For this we can use the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ , where we can put $x = a$ and $b = \sqrt {11} $ , this will give us:
$ \Rightarrow {x^3} + 11x = x\left( {{x^2} + 11} \right) = x\left( {x + \sqrt {11} } \right)\left( {x - \sqrt {11} } \right)$
Therefore, for the expression, ${x^3} + 11x$ we got the factors as $x,\left( {x - \sqrt {11} } \right){\text{ and }}\left( {x + \sqrt {11} } \right)$

Note:
In questions like this the use of algebraic identities plays a crucial role in the solution. An alternative approach to the same problem can be taken by using the method of inspection to find a value of ‘x’ for ${x^3} - 16x = 0$ , i.e. $x = 0{\text{ or }}x = 4$ . This gives us one of the factors as $\left( {x - 4} \right){\text{ or }}x$. Now divide the whole expression ${x^3} - 16x$ with $\left( {x - 4} \right){\text{ or }}x$ to get the other factor.