**Solve a quadratic equation by factorising :**

Solving a quadratic equations means, we have to find the value of the variable in the given quadratic equation.

A equation which is in the form of ax² + bx + c is known as quadratic equation.Here a,b and c are just numbers.Here a is not equal to zero.

Here we are going to see, how to factorise quadratic equation with detailed examples.

In a quadratic "Leading coefficient" means "coefficient of x²".

(i) If the coefficient is 1, we have to take the constant term and split it into two parts.

(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.

(iii) By grouping we get two factor in the (x+a) and (x+b)

(iv) Write each factors equal to zero and find the values of x.

(i) If it is not 1, then we have to multiply the coefficient of x² by the constant term and split it as two parts.

(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.

(iii) Simplify the factors by the coefficient of x² as much as possible.

(iv) Write each factors equal to zero and find the value of x.

Let us see some examples for better understanding.

**Example 1 :**

Solve x² + 17 x + 60 = 0

**Solution :**

**In the first step, we are going to check whether we have 1 as the coefficient of x² or not. **

**Since it is 1, we are going to take the last number. **

**Now let us find the factors of 60.**

Since the equation is in the form ax^{2} + bx + c, we have to take positive sign for both factors.

Here, 10 x 6 = 60 but 10 + 6 = 16 not 17 15 x 4 = 60 but 15 + 4 = 19 not 17 12 x 5 = 60 and 12 + 5 = 17 2 x 30 = 60 but 2 + 30 = 32 not 17 |

x² + 17 x + 60 = 0

(x + 12) (x + 5) = 0

x + 12 = 0 (or) x + 5 = 0

x = -12 (or) x = -5

**Example 2 :**

Solve x² - 14 x + 48 = 0

**Solution :**

**In the first step, we are going to check whether we have 1 as the coefficient of x² or not. **

**Since the middle term is negative, we have to take negative sign for both factors.**

x² - 14 x + 48 = 0

(x - 12) (x + 5) = 0

x + 12 = 0 (or) x + 5 = 0

x = -12 (or) x = -5

**Example 3 :**

Factor x² - x - 6

**Solution :**

**In the first step we are going to check whether we have 1 as the coefficient of x² or not. **

**Since thee middle and last terms are negative, we have to take negative sign for large number.**

**(x + 2) (x - 3)** are the factors of x² - x - 6.

**Example 4 :**

Factor x² + 2 x - 24

**Solution :**

**Since the last term is negative, we have to put negative sign for small number.**

**(x + 6) (x - 4)** are the factors of x² + 2x - 24.

**Example 5 :**

Factor 2 x² + x - 6

**Solution :**

To factor this quadratic equation, we have to multiply the coefficient of x² by the constant term

So we get -12. Now we have to split this -12 as the multiple of two numbers.

Since the last term is negative, we have to put negative sign for the least factor.

Now we have to divide the two numbers 4 and -3 by the coefficient of x² that is 2.If it is possible, we can simplify. Otherwise we have to write the numbers along with x.

So (x + 2) (2x - 3) are the factors of 2 x² + x - 6

**Example 6 :**

Factor 2 x² - 4 x - 16

**Solution :**

To factor this quadratic equation, we have to multiply the coefficient of x² by the constant term.

Now we have to split -32 as multiple of two numbers. For that we have to write the factors of 32.

Since the last and middle term are having negative sign, we have to put negative sign for the greater number.

Now we have to divide -8 and 4 by the coefficient of x² that is 2.If it is possible, we can simplify. Otherwise we have to write the numbers along with x.

So (x + 2) (x - 4) are the factors of 2 x² - 4 x - 16.

Let us see the the another example to understand the topic factoring quadratics when a is not equal to 1.

Let us see some examples on "Solve a quadratic equation by factorising".

**Example 7 :**

Factor 6 x² + 13 x - 5

**Solution :**

To factor this quadratic equation we have to multiply the coefficient of x² by the constant term.

If we multiply 6 and -5, we get -30.Now we have to split -30 as the product of two numbers.

The product of -2 and 15 is -30 and the simplified value is 13

So (3x - 1) (2x + 5) are the factors of 6 x² + 13 x - 5.

Let us see the the another example to understand the topic factoring quadratics when a is not equal to 1.

Let us see some examples on "Solve a quadratic equation by factorising".

**Example 8 :**

Factor 2 x² - 11 x + 12

**Solution :**

To factor this quadratic equation, we have to multiply the coefficient of x² by the constant term.

If we multiply 2 and 12, we get 24. Now we have to split 24 as the product of two numbers.

The product of -8 and -3 is 24 and the simplified value is -11.Since the middle term is having negative sign, we have to put negative sign for both 8 and 3.

(x - 4) and (2x - 3) are the factors of the quadratic equation 2 x² - 11 x + 12

- Factoring quadratic equations when the coefficient of x² is not 1
- Factoring using Algebraic identities
- Solving quadratic equation using completing the square method
- Solving quadratic equation by using quadratic formula
- Practical problem using quadratic equations

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