Solving Quadratic Equation by factorization has become challenging to some students. They face this difficulty probably because of the steps they use.

This method of providing solution to a quadratic equation is one of the simplest methods of solving quadratic equation.

After the lesson today, you will see with me that the method is not cumbersome to use in solving quadratic equations.

What you have to do is just follow the steps I will put down here. The fact that you have been taught by other tutors without understanding this method does not mean you can’t solve some mind-blogging problems on this topic.

I promise you, at the end of this lesson you are going to be thankful for opening this site. Let’s get familiarized with the steps required for solving problems using this method

** Step 1:** multiply the first term with the last terms

** Step 2:** Examine (Write down) the middle term

** Step 3:** Find two terms in x whose sum gives the middle term and the product gives the value in step 1

** Step 4:** Replace the middle term with the two factors and bracket them

** Step 5:** obtain the common terms and solve for two values of x

Now that you know the steps for solving any problem in quadratic equation, leit get started by solving some problems.

**Solutions to Quadratic Equation By Factorization**

__Example 1:__

Solve by factorization

9x^{2}-15x+4=0

Solution

The equation is 9x^{2}-15x+4=0

Step 1: the first term is 9x^{2} the last term is 4

9x^{2}x4= 36x^{2}

Step 2: the middle term is -15x

Step 3: two factors in x whose sum will give -15x and product (multiplication) will yield 36x^{2}

The factors are -12x and -3x because -12x x -3x= +36x^{2}

-x- = + and -12x + (-3x) = -12x-3x = -15x

Step 4: Replace -15x with the two factors =12x and -3x this because

9x^{2}-12x-3x+4=0

Bracket them

(9x^{2}-12x) – (3x+4) = 0

The common factor is 3x

3x(3x-4) – (3x-4) = 0

Since one factor appears twice,

Take one i.e (3x-4). The other factor is (3x-1)

Sir where did you get the 1 from ?

There is an invisible 1 that is with the minus sign

Therefore, the equation becomes

(3x-4) (3x-1) = 0

This means

3x-4=0 and 3×1=0

3x-4=0

Collect like terms

3x/3 =+4/3

x = +4/3

and

3x/3 = 1/3

x= 1/3

Therefore the two values of x are 4/3 or 1/3

Example 1:

Use factorization method to solve

3y^{2}=13y-10=0

Solution

The equation is 3y^{2}-13y-10=0

Step 1: the first term is 3y^{2} the last term is -10

That means 3y^{2}x-10=0 = 21-30y^{2}

Step 2: the middle term is -13y

Step 3: two factors whose sum gives the middle term that is -13y and the product gives -30y^{2}

The factors are -15y and 2y

Step 4: Replace the -13y with -15y and 2y

This gives 3y^{2}-15y+2y-10=0

We then bracket it

(3y^{2}-15y) + (2y-10)= 0

The common terms are 3y and 2

3y(y-5) + 2(y-5) = 0

Since we have (y-5) appearing twice, we pick one

(y-5) and (3y+2)

(y-5) (3y+2) = 0

That means

y-5=0 and 3y+2 =0

let’s now solve for the two values of y

y-5 = 0 3y+2 = 0

y = 5, y = -2/3

Therefore, the two values of y are 5 or -2/3

**Example 3:**

Solve by factorization

*x* +1/*x*= 2

Solution

Multiply through by *x*

*x*^{2} + 1 = 2_{x}

Take 2* _{x}* to the left hand side

*x*^{2} + *x*-2* _{x}* = 0

*x*^{2} -2* _{x}* + 1= 0

Step 1: The first term is *x*^{2} , the last term is 1

*x*^{2} x 1=*x*^{2}

Step 2: The middle term is ^{ }-2_{x}

Step 3: The two factors are –*x*and *-x*

*x*^{2} – *x*–*x*+1 = 0

we bracket it

(*x*^{2} – *x*) –(* x* -1) = 0

*x* (*x* – *x*) –(* x* -1) = 0

Have you observe the change in sign from the plus to minus?

This is because

-(* x*-1) = –* x*+1

That gives the previous term, I hope that is understood now! Ok then since (*x*-1) appear twice, we take 1.

(* x*-1) and (* x*-1)

(* x*-1) x * x*-1 = 0

* x* =1 and *x* =1

Therefore, the two values of *x* are 1 or 1 i.e 1 twice

Example 4:

Solve by factorization 6x^{2}-7x+2=0

Solution

As usual, we follow the steps.

The first term is 6x^{2} , the last term is 2

We multiply them

6x^{2}x2 = 12x^{2}

The middle term is -17x

The factors in x whose sum (addition) gives -7x and product will yield 12x^{2}

The two factors are

-4x and -3x

We replace -7x with the two factors

6x^{2}-4x-3x+2=0

We bracket them

(6x^{2}-4x – (3x+2)=0

2x(3x-2) – (3x-2) = 0

Since (3x-2) appear twice, we take one

(3x-2) and (2x-1)

(3x-2) = 0 2x-1 = 0

therefore, the values of x are 2/3 or ½

**Recommended:**

I strongly believe the difficulty you had when facing solution of quadratic equation by factorization has been erased.

One thing you have to understand in mathematics is that any topic you encounter has a particular rule or rules you need to follow.

That is why the solution of the quadratic equation by factorization is not an exception. Just abide by the steps have put down today, you will be able to solve any problem regarding quadratic.