What are the solutions to the equation x(2x+3)(4x^2-81)=0 ?

The solutions to the equation x(2x+3)(4x^2-81)=0 are x=0,x=-3/2 ,x=-9/2 ,x= 9/2

Find the zeros of x(2x+3)(4x^2-81)=0

The zeros of x(2x+3)(4x^2-81)=0 are x=0,x=-3/2 ,x=-9/2 ,x= 9/2

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x (2 x + 3) (4 x^{2} - 81) = 0

x = 0, x = - \frac{3}{2}, x = - \frac{9}{2}, x = \frac{9}{2}

Solution steps

x (2 x + 3) (4 x^{2} - 81) = 0

Factor

x (2 x + 3) (4 x^{2} - 81) : x (2 x + 3) (2 x + 9) (2 x - 9)

x (2 x + 3) (2 x + 9) (2 x - 9) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x + 3 = 0 or 2 x + 9 = 0 or 2 x - 9 = 0

Solve

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

Solve

2 x + 9 = 0 : x = - \frac{9}{2}

Solve

2 x - 9 = 0 : x = \frac{9}{2}

The solutions are

x = 0, x = - \frac{3}{2}, x = - \frac{9}{2}, x = \frac{9}{2}

x^{9} = 2

y^{4} + y^{3} + y^{2} = 0

x^{4} - 12 x^{2} + 4 = 0

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

a^{3} = - 64

What are the solutions to the equation x(2x+3)(4x^2-81)=0 ?
The solutions to the equation x(2x+3)(4x^2-81)=0 are x=0,x=-3/2 ,x=-9/2 ,x= 9/2
Find the zeros of x(2x+3)(4x^2-81)=0
The zeros of x(2x+3)(4x^2-81)=0 are x=0,x=-3/2 ,x=-9/2 ,x= 9/2