Factoring an equation implies getting the solution of the equation, by breaking the equation into such a simplified form, that it gives all the possible values of its variable. Solving equations by factoring generally involves equation having one variable only.

The procedure to solve an equation by factoring involves knowing the zero product property, which says that, if we have
two numbers or variables x and y, satisfying the relation x * y = 0,
then the result would be x = 0, or y = 0.**Equation with Degree One: **To
solve equation with one variable, we just need to simplify equation by
taking variable on one side of the sign and all the numbers on
the other side of sign and we will get the value of a variable.

Equation with Degree Two:

Factoring is done for equations having degree 2 or more. Now, an equation having degree 2, can be written as ax^{2} + bx
+ c = 0, where a, b, and c are numbers and a $\neq$ 0, because if a =
0, the equation gets reduced to the one degree equation. This form of an
equation is known as a quadratic equation.

- We will take all the
non zero terms on one side of the sign and with only zero on the other side.
- Then, we will use the zero product
property, wherein we will convert the equation in the form (x - c) (x -
d) = 0, where we will find two numbers c and d, which in addition must
be equal to b, and on multiplying must be equal to ac.
- The solution would be x = c and x = d, by zero product property.

Given below are some of the examples on solving equations by factoring.

Factor the equation, x

The given equation is in the form a^{2} + 2ab + b^{2}

Hence, we can write it as x^{2 }+ 2 (x) (2) + 2^{2} and we know (a + b) ^{2} = a^{2} + 2ab + b^{2}.

Thus, the solution will be (x + 2)^{2} or (x + 2) (x + 2).

Thus, the solution will be (x + 2)

Solve the equation by factoring 2x

To factorize this, we can take 2 common from all the terms.

Now, we get 2 (x^{2} +** 5x** + 4)

Now, we get 2 (x

= 2 [x^{2} + **4x + x** + 4]

This can be simplified to 2 [x (x + 4) + 1 (x + 4)]

This can be simplified to 2 [x (x + 4) + 1 (x + 4)]

= 2 (x + 4) (x + 1)

2 (x + 4) (x + 1) = 0

2 $\neq$ 0

So, (x + 4) (x + 1) = 0

By zero product property

x + 4 = 0 or x + 1 = 0

or x = - 4 and x = - 1

**Example 3:**

Factorize the equation x^{3} - 10x^{2} + 24x

**Solution:**

To factorize this, we take x common from all the terms.

Now, we get x (x^{2 }- 10x + 24)

Now, we can break the middle term as follows:

x (x^{2} - 6x - 4x + 24)

This can be simplified to x [x (x - 6) - 4 (x - 6)]

= x [(x - 6) (x - 4)]

= x (x - 6) (x - 4)

2 (x + 4) (x + 1) = 0

2 $\neq$ 0

So, (x + 4) (x + 1) = 0

By zero product property

x + 4 = 0 or x + 1 = 0

or x = - 4 and x = - 1

Factorize the equation x

To factorize this, we take x common from all the terms.

Now, we get x (x

Now, we can break the middle term as follows:

x (x

This can be simplified to x [x (x - 6) - 4 (x - 6)]

= x [(x - 6) (x - 4)]

= x (x - 6) (x - 4)

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