What are the solutions to the equation (4x-3)(x+2)(x-6)=0 ?

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

Find the zeros of (4x-3)(x+2)(x-6)=0

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

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(4 x - 3) (x + 2) (x - 6) = 0

x = \frac{3}{4}, x = - 2, x = 6

Decimal

x = 0.75, x = - 2, x = 6

Solution steps

(4 x - 3) (x + 2) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x - 3 = 0 or x + 2 = 0 or x - 6 = 0

Solve

4 x - 3 = 0 : x = \frac{3}{4}

Solve

x + 2 = 0 : x = - 2

Solve

x - 6 = 0 : x = 6

The solutions are

x = \frac{3}{4}, x = - 2, x = 6

25 x^{2} + 8 x^{3} - 441 = 0

x, 9 x^{3} + 9 y - 27 = 0

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

4 w^{4} + 40 w^{2} - 44 = 0

- a^{3} + 4 a^{2} - 5 a + 2 = 0

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