What are the solutions to the equation 16x^3-48x^2-25x+75=0 ?

The solutions to the equation 16x^3-48x^2-25x+75=0 are x=3,x=-5/4 ,x= 5/4

Find the zeros of 16x^3-48x^2-25x+75=0

The zeros of 16x^3-48x^2-25x+75=0 are x=3,x=-5/4 ,x= 5/4

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16 x^{3} - 48 x^{2} - 25 x + 75 = 0

x = 3, x = - \frac{5}{4}, x = \frac{5}{4}

Decimal Notation

x = 3, x = - 1.25, x = 1.25

Solution steps

16 x^{3} - 48 x^{2} - 25 x + 75 = 0

Factor

16 x^{3} - 48 x^{2} - 25 x + 75 : (x - 3) (4 x + 5) (4 x - 5)

(x - 3) (4 x + 5) (4 x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or 4 x + 5 = 0 or 4 x - 5 = 0

Solve

x - 3 = 0 : x = 3

Solve

4 x + 5 = 0 : x = - \frac{5}{4}

Solve

4 x - 5 = 0 : x = \frac{5}{4}

The solutions are

x = 3, x = - \frac{5}{4}, x = \frac{5}{4}

42 x^{5} - 210 x^{4} = 0

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

0 = x^{3} + 2 x - 1

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

2 x - 5 x^{2} + 8 x^{3} = 0

What are the solutions to the equation 16x^3-48x^2-25x+75=0 ?
The solutions to the equation 16x^3-48x^2-25x+75=0 are x=3,x=-5/4 ,x= 5/4
Find the zeros of 16x^3-48x^2-25x+75=0
The zeros of 16x^3-48x^2-25x+75=0 are x=3,x=-5/4 ,x= 5/4