Solve equations of the second degree by factoring

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

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