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

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

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

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

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x^{2} (x - 4) (2 x + 5) = 0

x = 0, x = 4, x = - \frac{5}{2}

Decimal Notation

x = 0, x = 4, x = - 2.5

Solution steps

x^{2} (x - 4) (2 x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 4 = 0 : x = 4

Solve

2 x + 5 = 0 : x = - \frac{5}{2}

The solutions are

x = 0, x = 4, x = - \frac{5}{2}

10 = \frac{π}{2} D^{3}

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

x^{4} = - 7 x^{3} + 16 x^{2} + 151 x + 230

m^{4} + 14 m^{3} + 54 m^{2} + 35 m - 6 = 0

x^{3} + 4 x^{2} - 15 x - 18 = 0

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