What are the solutions to the equation (x+2)(3x-15)(x^2-x-6)=0 ?

The solutions to the equation (x+2)(3x-15)(x^2-x-6)=0 are x=-2,x=5,x=3

Find the zeros of (x+2)(3x-15)(x^2-x-6)=0

The zeros of (x+2)(3x-15)(x^2-x-6)=0 are x=-2,x=5,x=3

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(x + 2) (3 x - 15) (x^{2} - x - 6) = 0

x = - 2, x = 5, x = 3

Solution steps

(x + 2) (3 x - 15) (x^{2} - x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or 3 x - 15 = 0 or x^{2} - x - 6 = 0

Solve

x + 2 = 0 : x = - 2

Solve

3 x - 15 = 0 : x = 5

Solve

x^{2} - x - 6 = 0 : x = 3, x = - 2

The solutions are

x = - 2, x = 5, x = 3

32 x - 16 = 2 x^{3} - x^{2}

(x + 5)^{3} = 0

x^{3} + 2 x + 5 = 0

15 x^{4} - 67 x^{2} + 72 = 0

x^{3} - 9 x^{2} - 9 x + 81 = 0

What are the solutions to the equation (x+2)(3x-15)(x^2-x-6)=0 ?
The solutions to the equation (x+2)(3x-15)(x^2-x-6)=0 are x=-2,x=5,x=3
Find the zeros of (x+2)(3x-15)(x^2-x-6)=0
The zeros of (x+2)(3x-15)(x^2-x-6)=0 are x=-2,x=5,x=3