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## Quadratic Equation

A quadratic equation is an expression written in the form **ax ^{2} + bx + c = 0** in which a, b, c are numerals and the power of x is 2. This expression can be factorized by using factorization of quadratic equation rules.

Note: The power of x can neither be fraction nor negative.

### Factoring Quadratic Equation

The process of solving quadratic equations using **factorization method **is a simple one.

I will state clearly in this article the easiest way to factorize quadratic equation. Happy learning.

### How Do I know if a quadratic equation can be Factorized?

To check if a quadratic equation can be factored, use the discriminant b^{2} – 4ac. If it gives a perfect square, it can be factorized but if it does not, it cannot be factorized.

We shall consider two cases in order to give you a clear picture of how simple it is to factorize.

**Case 1 for factorization of quadratic equation:** When the coefficient of x^{2} is 1 ( that is to say that a = 1)

The general quadratic equation now becomes

**x ^{2} + bx + c = 0**

### Steps to Factorize Quadratic Equation

- Check if the expression can be factored
- Check the sign of b and c
- If c is positive, get two factors of c whose product gives c and sum gives b. Both factors will have the sign of b.
- If c is negative, look for two factors of c whose product gives c and difference gives b. The numerically larger among the two factors will carry the sign of b and the smaller factor will carry the opposite sign.
- Replace the middle term with the factors
- Factorize by grouping the factors.

Example 1

x^{2} + 5x + 6 = 0

a = 1, b = 5, c = 6

Checking to know if the expression can be factored

b^{2} – 4ac = 5×5 – 4x1x6

= 25 – 24 = 1

SQRT (1) = 1. It gives a perfect square, so it can be factored.

Let’s move to the next step

You can see here that the coefficient of x^{2} is 1

The signs of c and b are positive

Possible pairs of factors of 6 are (1, 6) and (2, 3)

since c is positive, both factors will have the same sign as b and here b is positive. The factors whose product is 6 and sum yields b are (2, 3)

Replacing the middle term **5x** with the factors

Implies:

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

(x^{2} + 2x) (3x + 6)

x(x + 2) + 3(x+2)

(X+3)(x+2)

As simple as that.

**Case 2 for factorization of quadratic equation :** When the coefficient of x^{2} is not 1

In this case the general formula becomes

ax^{2} + bx + c = 0

Follow the step 1 above. After step 1, get the product of a and c = ac (the product must be taken as positive regardless of the sign).

write down all the possible pairs of factors of ac. If c is positive, get two factors of ac whose product gives ac and sum gives b. Both factors should have the sign of b.

If c is negative, look for two factors of ac whose product gives ac and difference gives b. The numerically larger among the two factors will carry the sign of b and the smaller factor will carry the opposite sign.

Then, follow step 5 to 6.

Example 2

Factorize 6x^{2} + 11x + 3

Let’s get working.

a = 6, b = 11, c = 3

ac = 6×3 = 18

Possible pairs of factors of 18 are (1,18), (2, 9) and (3, 6)

c is positive.

Therefore, the two factors will carry the sign of b and their sum should give b

2+9 = 11.

This implies that 2 and 9 are the pairs of factors that satisfied the condition.

Let’s replace the middle term **’11x’** with the two factors

6x^{2} + 2x + 9x + 3 = 0

factoring by grouping

( 6x^{2} + 2x )( 9x + 3 )

Taking out the common factors

2x(3x + 1) + 3(3x+1)

(2x + 3)(3x+1)

That is it.

I hope you enjoyed it. Please use the comment section if you have any question and remember to subscribe for more lessons.

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Read Also:

**Simultaneous Equations: Solved Examples and Shortcut Tricks**

Last Updated on July 8, 2019 by Admin

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