Solve equations of the second degree by factoring could be several times the most simple and fastest solve the quadratic equation as long as you know how factor.
I urge you to study or review the following important unit: factoring
Example # 1: Solving the equations of the second degree by factoring
Solve x 2 + 3 x + 2 = 0.
First, you need to factor x 2 + 3 x + 2
Given that the coefficient of x 2 is 1 (x 2 = 1x 2), you can factor in looking for factors of the last Parliament (latter term is 2) which adds to the coefficient of the second term (3 x, factor is 3)
2 = 1 × 2
-2 =-1 × 2
Given that 1 + 2 = 3, and 3 is the coefficient of the second term, x 2 + 3 x + 2 = (x + 2) × (x + 1)
x 2 + 3 x + 2 = 0 gives:
(x + 2) × (x + 1) = 0
(x + 2) × (x + 1) = 0 when x + 2 = 0 or x + 1 = 0.
x + 2 = 0 when x = – 2
x + 1 = 0 when x = – 1
Let us now check x =-2 and x =-1 are solutions of 2 x + 3 x + 2 = 0
(-2) 2 + 3 ×-2 + 2 = 4 + 6 + 2 = 3 + 6 = 0
(-1) 2 + 3 ×-1 + 2 = 1 + 3 + 2 = 3 + 3 = 0
If instead you were resolution x-2 + 3 x + 2 = 0, you’ll:
x-2 + 3 x + 2 = 0.
( x + -2) × ( x + -1) = 0
-(x + 2) × (x +-1) = 0 when x +-2 = 0 or x +-1 = 0.
-x + 2 = 0 when x = 2
-x + 1 = 0 when x = 1
Make sure that they are indeed solutions
Example # 2: Resolution of the equations of the second degree by factoring
Solve x 2 + x-30 = 0.
First, you need to factor x 2 + x – 30
-30 = 30 ×-1
-30-15 × 2 =
-30 = 6 ×-5
-30 =-30 × 1
-30-15 × 2 =
-30 =-6 × 5
Since only 6 +-5 = 1 and 1 are the coefficient of the second term (x = 1x), x 2 + x – 30 = (x + 6) × (x – 5)
x 2 + x-30 = 0 gives:
(x + 6) × (x – 5) = 0
(x + 6) × (x – 5) = 0 when x + 6 = 0 or x – 5 = 0.
x + 6 = 0 when x = – 6
x – 5 = 0 when x = 5
Let us now check x =-6 and x = 5 are indeed x 2 + x-30 = 0 solutions
(-6) 2 +-6 – 30 = 36-6-30 = 30-30 = 0
2 (5) + 5 + 5-30-30 = 25 = 30-30 = 0.
If instead you were resolution x 2 + x – 30 = 0, you’ll:
x 2 + – x – 30 = 0.
(x – 6) × (x + 5) = 0
(x – 6) × (x + 5) = 0 when x – 6 = 0 or x + 5 = 0.
x – 6 = 0 when x = 6
x + 5 = 0 when x = – 5
Make sure that they are indeed solutions
Solve equations of the second degree by factoring can become very difficult. See below:
Example # 3: Solving the equations of the second degree by factoring
6 x 2 + 27 x + 30 = 0.
First factor 6 x 2 + 27 x + 30
6 x 2 + x + 30 = 27 (3 x +?) × (2 x +?) or (x 6 +?) × (x +?)
Now, the last factor term 30
30 = 30 × 1
30 = 15 × 2
30 = 6 × 5
To obtain 27 x, you should try to multiply cross below. There are 6 of them. Cross multiply and add!
6 x x
30 1
x 6 + 30 x = x 36
6 x x
15 2
x 12 + 15 x = x 27
6 x x
6 5
x 30 + 6 x = x 36
3 x 2 x
30 1
3 x + x 60 = x 63
3 x 2 x
15 2
x 6 + 30 x = x 36
3 x 2 x
6 5
15 x + 12 x = x 27
You have two choices in bold!
6 x 2 + 27 x + 30 = 0.
(6 x + 15) × (x + 2) = 0
x 6 + 15 = 0.
x 6 = – 15
x = – 15/6
x + 2 = 0.
x = – 2