**Solving quadratic equations by factoring examples :**

Here we are going to see some example problems on solving quadratic equations using the method factoring.

Whenever we have a quadratic equation in the form ax² + bx + c and we need to factor this, first we have to check whether the coefficient of x² is 1 or not.

- If it is 1, then we have to take the constant term and split it into two factors.
- In which the product of two factors must be equal to the constant term and the simplified value of those factors equal to the middle term, that is coefficient of x.
- Write each factors is in the form (x + a) (x + b) according to the sign.
- Set each factors equal to zero to get the value of x.

- Multiply the coefficient of x² and constant term.
- Split this product into two factors such that their sum is equal to the coefficient of x .
- Divide each factors by the coefficient of x².
- If it is possible we can simplify, otherwise we have to write the denominator along with x.
- write each factors in the form (x + a) (x + b).
- Set each factors equal to zero to get the value of x.

**Example 1 :**

Solve x² + 9 x + 14 = 0

**Solution :**

**Since the coefficient of x**² is 1, split the constant term that into two parts.

14 = 2 ⋅ 7, 2 + 7 = 9

x² + 9 x + 14 = 0

(x + 2) (x + 7) = 0

x - 2 = 0 x + 7 = 0

x = 2, x = -7

Hence the solution is {2, -7}

Let us look into the next example problem on "Solving quadratic equations by factoring examples".

**Example**** 2 :**

Solve x² - 9 x + 14 = 0

**Solution :**

**Since the coefficient of x**² is 1, split the constant term that into two parts.

14 = -2 ⋅ (-7) , -2 - 7 = -9

x² - 9 x + 14 = (x - 2) (x - 7)

(x - 2) (x - 7) = 0

x - 2 = 0 x - 7 = 0

x = 2, x = 7

Hence the solution is {2, 7}

Let us look into the next example problem on "Solving quadratic equations by factoring examples".

**Example 3 :**

Solve x² + 2 x - 15 = 0

**Solution :**

**Since the coefficient of x**² is 1, split the constant term that into two parts.

-15 = -3 ⋅ 5 , -3 + 5 = 2

x² + 2 x - 15 = (x - 3) (x + 5)

(x - 3) (x + 5) = 0

x - 3 = 0 x + 5 = 0

x = 3, x = -5

Hence the solution is {3, -5}

**Example 4 :**

Solve x² - 2 x - 15 = 0

**Solution :**

**Since the coefficient of x**² is 1, split the constant term that into two parts.

-15 = -5 ⋅ 3 , -5 + 3 = -2

x² - 2 x - 15 = (x - 5) (x + 3)

(x - 5) (x + 3) = 0

x - 5 = 0 x + 3 = 0

x = 5, x = -3

Hence the solution is {-3, 5}

**Example 5 :**

Solve 2x² + 15 x + 27 = 0

**Solution :**

**Since the coefficient of x**² is not 1, multiply the coefficient of **x**² by the constant term and split it into two parts.

2 ⋅ 27 = 54

54 = 6 ⋅ 9 , 6 + 9 = 15

(x + 3) (2x + 9) = 0

x + 3 = 0 2x + 9 = 0

x = -3, x = -9/2

Hence the solution is {-3, -9/2}

**Example 6 :**

Solve 2x² - 15 x + 27 = 0

**Solution :**

**Since the coefficient of x**² is not 1, multiply the coefficient of **x**² by the constant term and split it into two parts.

2 ⋅ 27 = 54

54 = -6 ⋅ (-9) , -6 - 9 = -15

2x² - 6 x - 9x + 27 = 0

Factor 2x from the first two terms and factor 9 from the third and fourth terms

2x (x - 3) - 9(x - 3) = 0

(2x - 9) (x - 3) = 0

2x - 9 = 0, x - 3 = 0

2x = 9

x = 9/2 and x = 3

Hence the solution is {9/2, 3)

**Example 7 :**

Solve 2x² + 15 x - 27 = 0

**Solution :**

**Since the coefficient of x**² is not 1, multiply the coefficient of **x**² by the constant term and split it into two parts.

2 ⋅ (-27) = -54

-54 = 18 ⋅ (-3) , 18 - 3 = 15

2x² + 18 x - 3 x - 27 = 0

Factor 2x from the first two terms and factor 3 from the third and fourth terms

2x (x - 9) - 3(x - 9) = 0

(2x - 3) (x - 9) = 0

2x - 3 = 0, x - 9 = 0

2x = 9

x = 9/2 and x = 3

Hence the solution is {9/2, 3)

After having gone through the stuff given above, we hope that the students would have understood "Solving quadratic equations by factoring examples".

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