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

Factoring quadratics with a leading coefficient of 1

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

(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 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.

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 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.

**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

- Solving quadratic equation by completing the square
- Solving quadratic equation by sum and product of roots
- Solving quadratic equation by formula
- Solving quadratic inequalities graphically
- Solving quadratic inequalities algebraically
- Solving word problems using quadratic equation

HTML Comment Box is loading comments...

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**