What are the solutions to the equation (3x-2)(x-4)(x+5)=0 ?

The solutions to the equation (3x-2)(x-4)(x+5)=0 are x= 2/3 ,x=4,x=-5

Find the zeros of (3x-2)(x-4)(x+5)=0

The zeros of (3x-2)(x-4)(x+5)=0 are x= 2/3 ,x=4,x=-5

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(3 x - 2) (x - 4) (x + 5) = 0

x = \frac{2}{3}, x = 4, x = - 5

Solution steps

(3 x - 2) (x - 4) (x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 2 = 0 or x - 4 = 0 or x + 5 = 0

Solve

3 x - 2 = 0 : x = \frac{2}{3}

Solve

x - 4 = 0 : x = 4

Solve

x + 5 = 0 : x = - 5

The solutions are

x = \frac{2}{3}, x = 4, x = - 5

10 x^{2} - 15 = 3 x (6 x^{2} - 5 x)

1 = x^{5}

x^{4} + 8 x^{3} + 17 x^{2} - 2 x - 24 = 0

3 m^{3} - 12 m = 0

x^{3} - 4 x^{2} + 19 x + 5 = 0

What are the solutions to the equation (3x-2)(x-4)(x+5)=0 ?
The solutions to the equation (3x-2)(x-4)(x+5)=0 are x= 2/3 ,x=4,x=-5
Find the zeros of (3x-2)(x-4)(x+5)=0
The zeros of (3x-2)(x-4)(x+5)=0 are x= 2/3 ,x=4,x=-5