## SOLVING QUADRATIC EQUATIONS BY FACTORING

On this webpage,"Solving quadratic equations by factoring" we are going to see how to solve quadratic equation in a simple way.

Generally we have two types of equations

(i) If the coefficient is 1 we have to take the constant term and we have to 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) Now we have to write these numbers in the form of (x + a) and (x +b)

(iv) By equating the factors equal to zero.We can find the value of x.  solving quadratic equations by factoring

## Example problem of solving quadratic equations by factoring

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. That is 60 and we are going to factors of 60.

All terms are having positive sign. So we have to put 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

(x + 12) (x + 5) are the factors

x + 12 = 0        x + 5 = 0

x = -12            x = -5

Solution is {-12,-5}

This is just example of solving quadratic equations by factoring.If you need more example problems of solving quadratic equations by factoring please click the below link.

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Factoring quadratic equations when a isn't 1

(i) If it is not 1 then we have to multiply the coefficient of x² by the constant term and we have to 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) Divide the factors by the coefficient of x². Simplify the factors by the coefficient of x² as much as possible.

(iv) Write the remaining number along with x.

## Example problem of solving quadratic equations by factoring

Example 2:

Factor 2 x² + x - 6

Solution:

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

So that we get -12, now we have to split -12 as the multiple of two numbers.

Since the last term is having negative sign.So we have to put negative sign for the least number.

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.

(x + 2) (2x - 3)  = 0

x + 2 = 0         2 x - 3 = 0

x = -2             2 x = 3

x = 3/2

This is just one example problem to show solving quadratic equations by factoring. If you want more example please click the below link.

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## More example problems

Problem 1:

Solve by factoring method (2x + 3)²  - 81 = 0

Solution:

(2x + 3)²  - 81 = 0

Now we are going to compare (2x + 3)² with the algebraic identity (a+b)² = a² + 2 a b + b²

(2x + 3)²  - 81 = 0

(2 x)² + 2 (2 x) (3) + 3² - 81 = 0

4 x² + 12 x + 9 – 81 = 0

4 x² + 12 x – 72 = 0

Now we are going to divide the whole equation by 4

x² + 3 x – 18 = 0

(x + 6) (x – 3) = 0

x + 6 = 0      x – 3 = 0

x = -6             x = 3

Alternate way:

(2x + 3)²  - 81 = 0

(2x + 3)²  = 81

(2 x + 3) = √81

(2 x + 3) = ± 9

2 x + 3 = 9                           2 x + 3 = - 9

2 x = 9 – 3                           2 x = - 9 – 3

2 x = 6                                  2 x = -12

x = 6/2                                   x = -12/2

x = 3                                        x = -6

Verification:

(2x + 3)²  - 81 = 0

if x = 3

(2 (3) + 3)²  - 81 = 0

(6 + 3)²  - 81 = 0

9²  - 81 = 0

81 - 81 = 0

0 = 0

if x = -6

(2 (-6) + 3)²  - 81 = 0

(-12 + 3)²  - 81 = 0

(-9)²  - 81 = 0

81 - 81 = 0

0 = 0

Now let us see the next example problem of the concept "solving quadratic equations by factoring".

Problem 2:

Solve by factoring method 3 x² – 5 x – 12 = 0

Solution:

3 x² – 5 x – 12 = 0

3 x² – 9 x  + 4 x – 12 = 0

3 x (x – 3) + 4 (x – 3) = 0

(3 x +  4) (x - 3) = 0

3 x + 4 = 0                 x – 3 = 0

3 x = -4                          x = 3

x = -4/3

Verification:

3 x² – 5 x – 12 = 0

if x = 3

3 (3)² – 5 (3) – 12 = 0

3(9) - 15 - 12 = 0

27 - 27 = 0

0 = 0

if x = -4/3

3 (-4/3)² – 5 (-4/3) – 12 = 0

(16/3) + (20/3) - (36/3) = 0

(16 + 20 - 36)/3 = 0

(36 - 36)//3 = 0

0/3 = 0

0 = 0

Now let us see the next example problem of the concept "solving quadratic equations by factoring".

Problem 3:

Solve by factoring method √5 x² + 2 x – 3√5 = 0

Solution:

√5 x² + 5 x  - 3 x – 3√5 = 0

√5 x (x + √5) – 3 (x + √5) = 0

(√5 x – 3) (x + √5) = 0

√5 x – 3 = 0                     x + √5 = 0

√5 x = 3                                x = - √5

x = 3/√5

Verification:

√5 x² + 2 x – 3√5 = 0

if x = 3/√5

√5 (3/√5)² + 2 (3/√5) – 3√5 = 0

√5(9/5) + (6/√5) – 3√5 = 0

(9 √5/5) + (6√5/5) - (15√5)/5 = 0

(9 √5+ 6√5 - 15√5)/5 = 0

(15 √5 - 15 √5)/5  = 0

0/5 = 0

0 = 0

Now let us see the next example problem of the concept "solving quadratic equations by factoring".

Problem 4:

Solve by factoring method 3 (x² – 6) = x (x + 7) – 3

Solution:

3 x² – 18 = x² + 7 x – 3

3x² - x² – 7 x – 18 + 3 = 0

2 x² – 7 x – 15 = 0

2 x² – 10 x + 3 x – 15 = 0

2 x (x – 5) + 3( x – 5) = 0

(2 x + 3) (x – 5) = 0

2 x + 3 = 0               x – 5 = 0

2x = -3                       x = 5

x = -3/2

Verification:

2 x² – 7 x – 15 = 0

if x = -3/2

2 (-3/2)² – 7(-3/2) – 15 = 0

2 (9/4) + 21/2 – 15 = 0

(9/2) + (21/2)  - 15 = 0

(9 + 21 - 30)/2 = 0

(30 - 30)/2 = 0

0 = 0

if x = 5

2 x² – 7 x – 15 = 0

2 (5)² - 7(5) - 15 = 0

2 (25) - 35 - 15 = 0

50 - 50 = 0

0 = 0

Now let us see the next example problem of the concept "solving quadratic equations by factoring".

Problem 5:

Solve by factoring method 3 x – (8/x) = 2

Solution:

(3 x² – 8)/x = 2

(3 x² – 8) =2 x

3 x² - 2 x – 8 = 0

3 x² - 6 x + 4 x – 8 = 0

3 x (x – 2) + 4 (x – 2) = 0

(3 x + 4) (x – 2) = 0

3 x + 4 = 0                 x – 2 = 0

3x = -4                       x = 2

x = -4/3

Verification:

3 x² - 2 x – 8 = 0

if x = -4/3

3 (-4/3)² - 2(-4/3) – 8 = 0

(16/3) + (8/3) - 8 = 0

(16 + 8 - 24)/3 = 0

(24 - 24)/3 = 0

0 = 0

if x = 2

3 (2)² - 2 (2) – 8 = 0

3 (4) - 4 - 8 = 0

12 - 4 - 8 = 0

12 - 12 = 0

0 = 0

Now let us see the next example problem of the concept "solving quadratic equations by factoring".

Problem 6:

x + (1/x) = (26/5)

Solution:

(x² + 1)/x = 26/5

5 (x² + 1) = 26 x

5 x² + 5 = 26 x

5 x² - 26 x + 5 = 0

5 x² - 25 x – x  + 5 = 0

5 x (x -5) – 1(x – 5) = 0

(5 x – 1) (x – 5) = 0

5 x – 1 = 0             x – 5 =0

5 x = 1                   x = 5

x = 1/5

Verification:

5 x² - 26 x + 5 = 0

if x = 1/5

5(1/5)² - 26(1/5) + 5 = 0

(1/5) - (26/5) + 5 = 0

(1 - 26 + 25)/5 = 0

(26 - 26)/5 = 0

0/5 = 0

0 = 0

if x = 5

5 (5)² - 26(5) + 5 = 0

5 (25) - 130 + 5 = 0

125 + 5 - 130 = 0

130 - 130 = 0

0 = 0

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