What are the solutions to the equation (x^2-4)(x+3)=x^2+5x+6 ?

The solutions to the equation (x^2-4)(x+3)=x^2+5x+6 are x=-2,x=-3,x=3

Find the zeros of (x^2-4)(x+3)=x^2+5x+6

The zeros of (x^2-4)(x+3)=x^2+5x+6 are x=-2,x=-3,x=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^{2} - 4) (x + 3) = x^{2} + 5 x + 6

x = - 2, x = - 3, x = 3

Solution steps

(x^{2} - 4) (x + 3) = x^{2} + 5 x + 6

Expand

(x^{2} - 4) (x + 3) : x^{3} + 3 x^{2} - 4 x - 12

x^{3} + 3 x^{2} - 4 x - 12 = x^{2} + 5 x + 6

Move

6

to the left side

x^{3} + 3 x^{2} - 4 x - 18 = x^{2} + 5 x

Move

5 x

to the left side

x^{3} + 3 x^{2} - 9 x - 18 = x^{2}

Move

x^{2}

to the left side

x^{3} + 2 x^{2} - 9 x - 18 = 0

Factor

x^{3} + 2 x^{2} - 9 x - 18 : (x + 2) (x + 3) (x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x + 3 = 0 or x - 3 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

The solutions are

x = - 2, x = - 3, x = 3

7 x^{4} = 20 x^{2}

- 4 x^{3} + 3 x^{2} + 2 x = 0

0 = 4 x^{3} - 4

v^{3} - 4 v = 0

4 x^{3} - 256 x = 0

What are the solutions to the equation (x^2-4)(x+3)=x^2+5x+6 ?
The solutions to the equation (x^2-4)(x+3)=x^2+5x+6 are x=-2,x=-3,x=3
Find the zeros of (x^2-4)(x+3)=x^2+5x+6
The zeros of (x^2-4)(x+3)=x^2+5x+6 are x=-2,x=-3,x=3