Question

How do you factor the equation ${x^2} - 4x = 0$ ?

Answer 1

Hint: Factoring is the writing of a number or another mathematical object as a product of several factors. Solve the equation by taking $x$ as common from the whole equation and then make them equal to 0 which will give the two values of $x$.

Complete Step by Step Solution:
Factorization or factoring in mathematics, is the writing of a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, the factorization of 15 is $3 \times 5$ and factorization of the polynomial ${x^2} - 4$ is $\left( {x - 2} \right)\left( {x + 2} \right)$.
The equation given in the question is ${x^2} - 4x = 0$.
$ \Rightarrow {x^2} - 4x = 0$
Taking $x$ common from the whole equation, we get –
$ \Rightarrow x\left( {x - 4} \right) = 0 \cdots \left( 1 \right)$
Case – 1: The equation (1) can be written as –
$
   \Rightarrow x = \dfrac{0}{{x - 4}} \\
   \Rightarrow x = 0 \\
 $
Therefore, we got one value of $x$ equal to 0.
Case – 2: Equation (1) can also be written in the form of –
$
   \Rightarrow x - 4 = \dfrac{0}{x} \\
   \Rightarrow x - 4 = 0 \\
   \Rightarrow x = 4 \\
 $
Therefore, another value for $x$ is equal to 4.
So, we get the two values of $x$ which are 0 and 4.
To check whether the founded value of $x$ is correct or not we will put these values one by one in the equation –
Case – 1: Putting $x = 0$ in the equation ${x^2} - 4x = 0$ , we get –
$
   \Rightarrow {0^2} - 4 \times 0 = 0 \\
   \Rightarrow 0 - 0 = 0 \\
 $
L.H.S. = R.H.S.
So, $x = 0$ satisfies the equation ${x^2} - 4x = 0$. Hence, this value of $x$ is correct.
Case – 2: Putting $x = 4$ in the equation ${x^2} - 4x = 0$, we get –
$
   \Rightarrow {4^2} - 4 \times 4 = 0 \\
   \Rightarrow 16 - 16 = 0 \\
 $
L.H.S. = R.H.S.
So, $x = 4$ satisfies the equation ${x^2} - 4x = 0$. Hence, this value of $x$ is also correct.

Hence, both the values of $x$ are correct and satisfies the equation.

Note: Many students can make mistakes and can find only one value for $x$, but it is not the correct way of factoring the equation. Try to find the values of $x$ as much as possible and check whether they will satisfy the equation according to the question.