What are the solutions to the equation 4(3-3x)(8-2x^2)=0 ?

The solutions to the equation 4(3-3x)(8-2x^2)=0 are x=1,x=-2,x=2

Find the zeros of 4(3-3x)(8-2x^2)=0

The zeros of 4(3-3x)(8-2x^2)=0 are x=1,x=-2,x=2

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4 (3 - 3 x) (8 - 2 x^{2}) = 0

x = 1, x = - 2, x = 2

Solution steps

4 (3 - 3 x) (8 - 2 x^{2}) = 0

Factor

4 (3 - 3 x) (8 - 2 x^{2}) : 24 (x - 1) (x + 2) (x - 2)

24 (x - 1) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 2 = 0 or x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = 1, x = - 2, x = 2

7 t^{3} + 2 t^{2} - t = 0

12 x^{3} + 12 x^{2} - 24 x = 0

x^{3} + 5.4 x^{2} - 52.1 = 0

x^{3} - 15 x^{2} + 50 x = 0

(x + 1)^{4} = 16 (x - 1)^{4}

What are the solutions to the equation 4(3-3x)(8-2x^2)=0 ?
The solutions to the equation 4(3-3x)(8-2x^2)=0 are x=1,x=-2,x=2
Find the zeros of 4(3-3x)(8-2x^2)=0
The zeros of 4(3-3x)(8-2x^2)=0 are x=1,x=-2,x=2