What are the solutions to the equation (x^2-x-6)*(x+4)=0 ?

The solutions to the equation (x^2-x-6)*(x+4)=0 are x=-2,x=3,x=-4

Find the zeros of (x^2-x-6)*(x+4)=0

The zeros of (x^2-x-6)*(x+4)=0 are x=-2,x=3,x=-4

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(x^{2} - x - 6) \cdot (x + 4) = 0

x = - 2, x = 3, x = - 4

Solution steps

(x^{2} - x - 6) (x + 4) = 0

Factor

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

(x + 2) (x - 3) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 3 = 0 or x + 4 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 3 = 0 : x = 3

Solve

x + 4 = 0 : x = - 4

The solutions are

x = - 2, x = 3, x = - 4

8 x^{3} - 8 x = 0

6 x = 2 x^{3} + 11 x^{2}

x^{4} - 5 x + 4 = 0

s^{4} - 6 s - 15 s = 3

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

What are the solutions to the equation (x^2-x-6)*(x+4)=0 ?
The solutions to the equation (x^2-x-6)*(x+4)=0 are x=-2,x=3,x=-4
Find the zeros of (x^2-x-6)*(x+4)=0
The zeros of (x^2-x-6)*(x+4)=0 are x=-2,x=3,x=-4