# Solving Quadratic equations using factorization method

In this chapter we will learn methods to solve quadratic equations with examples.

A the end of the chapter, problems are also given for your practice.

## Solving Quadratic equation

There are two methods for solving quadratic equation;

(a) Factorization method

(b) Formula method

Here we will discuss the factorization method.

### Factorization method for quadratic equation

In this method we divide the given quadrilateral into small components.

We will learn the method using example.

Consider the below quadratic equation;

\mathtt{x^{2} -6x+5=0}

To solve the equation follow the below steps;

(a) Multiply coefficient of \mathtt{x^{2}} and constant.

\mathtt{x^{2} -6x+5=0}\\\ \\ \mathtt{\Longrightarrow \ 1\times 5}\\\ \\ \mathtt{\Longrightarrow \ 5}

(b) Identify the coefficient of x

Here coefficient of x is 6.

(c) Now you have to identify two numbers which on multiplication produce 5 and on addition produce 6.

If a & b are two numbers then;

a . b = 5
a + b = 6

The value of a & b can be found by factorizing number 5 into smaller components.

Different possibilities of number 5 are;
5 = 5 x 1

Note that 5 & 1 are the numbers which of multiplication produce 5 and on addition produce 6.

Hence, 5 & 1 are the required numbers.

(d) Break the coefficient of x into smaller components using number 5 & 1.

\mathtt{x^{2} -6x+5=0}\\\ \\ \mathtt{x^{2} -( 1+5) x+5=0}

Now do the factorization.

\mathtt{x^{2} -x-5x+5=0}\\\ \\ \mathtt{x( x-1) -5( x-1) =0}\\\ \\ \mathtt{( x-5)( x-1) =0}\\\ \\ \mathtt{x=5\ \ \ \&\ \ x\ =1}

Hence, x= 1 and x = 5 are the solution of given quadratic equations.

The graph of quadratic equation is shown below;

I hope you understood the above process.

Note:
Not all quadratic equation will get solved using factorization method.

To solve complex quadratic equation problems, formula method is the best bet. We have discussed formula method in next chapter.

Let us some problems related to above concept.

## Quadratic equation factorization – Solved problems

Example 01
Solve the quadratic equation

\mathtt{x^{2} -25x+100=0}

Solution
Follow the below steps;

(a) Multiply coefficient of \mathtt{x^{2}} and constant

⟹ 1 x 100

⟹ 100

(b) Identify the coefficient of x.

Number 25 is the coefficient of x.

(c) Now find two numbers which on multiplication produce 100 and on addition/subtraction produce 25.

To find the numbers factorize 100.

100 = 20 x 5

Here 20 and 5 are the two numbers which fulfill the criteria.

(d) Break the coefficient x into smaller component and then factorize.

\mathtt{x^{2} -25x+100=0}\\\ \\ \mathtt{x^{2} -( 5+20) x+100=0}\\\ \\ \mathtt{x^{2} -5x-20x+100=0}\\\ \\ \mathtt{x( x-5) -20( x-5) =0}\\\ \\ \mathtt{( x-20)( x-5) =0}

x= 20 & x = 5

Hence, x=20 and x = 5 are the solution of quadratic equation.

Example 02
Solve the quadratic equation

\mathtt{x^{2} -4x-45=0}

Solution
(a) Multiply coefficient of \mathtt{x^{2}} and constant

⟹1 x 45

⟹ 45

(b) Coefficient of x is 4

(c) Find two numbers a & b which on multiplication produce 45 and on addition/subtraction produce 4.

Factorizing 45

Here 9 & 5 are the two numbers

Multiplication
45 = 9 x 5

Subtraction
9 – 5 = 4

(d) Break the coefficient of x using these two numbers and then factorize

\mathtt{x^{2} -4x-45=0}\\\ \\ \mathtt{x^{2} -( 9-5) x-45=0}\\\ \\ \mathtt{x^{2} -9x+5x-45=0}\\\ \\ \mathtt{x( x-9) +5( x-9) =0}\\\ \\ \mathtt{( x+5)( x-9) =0}\\\ \\ \mathtt{x=-5\ and\ x=9}

Hence, x = -5 and x = 9 are the solution.

Example 03
Find solution of below quadratic equation

\mathtt{x^{2} +10x+25=0}

Solution
(a) Multiply coefficient of \mathtt{x^{2}} and constant.

⟹ 1 x 25

⟹ 25

(b) Coefficient of x is 10

(c) Find two numbers a & b which on multiplication produce 25 and on addition/subtraction produce 10.

Factorizing 25

25 = 5 x 5

The two numbers are 5 & 5.

(d) Break the coefficient of x into smaller components using above two numbers.

\mathtt{x^{2} +10x+25=0}\\\ \\ \mathtt{x^{2} +( 5+5) x+25=0}\\\ \\ \mathtt{x^{2} +5x+5x+25=0}\\\ \\ \mathtt{x( x+5) +5( x+5) =0}\\\ \\ \mathtt{( x+5)( x+5) =0}\\\ \\ \mathtt{( x+5)^{2} =0}\\\ \\ \mathtt{x=-5}

Hence, x = -5 is the solution.

Example 04
Solve the quadratic equation.

\mathtt{3x^{2} +11x+6=0}

Solution
(a) Multiply coefficient of \mathtt{x^{2}} and constant

⟹ 3 x 6

⟹ 18

(b) The coefficient of x is 11

(c) Find two number which on multiplication produce 18 and on addition/subtraction produce 11.

Factorize 18

The two numbers are 2 and 9.

Multiplication
⟹ 2 x 9
⟹ 18

⟹ 2 + 9
⟹ 11

(d) Factorize the quadratic equation using these two numbers.

\mathtt{3x^{2} +11x+6=0}\\\ \\ \mathtt{3x^{2} +( 9+2) x+6=0}\\\ \\ \mathtt{3x^{2} +9x+2x+6=0}\\\ \\ \mathtt{3x( x+3) +2( x+3) =0}\\\ \\ \mathtt{( 3x+2) \ ( x+3) =\ 0}\\\ \\ \mathtt{3x+2\ =0\ \ \ \&\ x+3=0}\\\ \\ \mathtt{x=\frac{-2}{3} \ \ \&\ x=-3}

Hence, x =-2/3 and x=-3 is the solution.

Example 05

\mathtt{4x^{2} +4x-24=0}

Solution
(a) Multiply coefficient of \mathtt{x^{2}} and constant

⟹ 4 x 24

⟹ 96

(b) The coefficient of x is 4

(c) Find two number which on multiplication produce 96 and addition/subtraction produce 4.

Factorize 96.

The two possible numbers are 12 and 8.

Multiplication test
⟹ 12 x 8
⟹ 96

Addition / Subtraction test
⟹ 12 – 8
⟹ 4

Hence, numbers 12 and 8 are the right choice.

(d) Factorize the quadratic equation.

\mathtt{4x^{2} +4x-24=0}\\\ \\ \mathtt{4x^{2} +( 12-8) x-24=\ 0}\\\ \\ \mathtt{4x^{2} +12x-8x-24=0}\\\ \\ \mathtt{4x( x+3) -8( x+3) =0}\\\ \\ \mathtt{( 4x-8)( x+3) =0}\\\ \\ \mathtt{4x-8=0\ \ \&\ \ x\ +3\ =\ 0}\\\ \\ \mathtt{x=2\ \&\ \ x=-3}

Hence, x = 2 and x = -3 is the solution.

Next chapter : Solving quadratic equation using formula

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