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

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

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

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

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

x = 0, x = 3, x = \frac{5}{2}

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

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

The solutions are

x = 0, x = 3, x = \frac{5}{2}

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

x^{3} + x^{2} - 7 x + 15 = 0

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

x^{2} (x - 1) (x + 2) = 0

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

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