What are the solutions to the equation x^3+60=23x+2x^2 ?

The solutions to the equation x^3+60=23x+2x^2 are x=3,x=4,x=-5

Find the zeros of x^3+60=23x+2x^2

The zeros of x^3+60=23x+2x^2 are x=3,x=4,x=-5

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x^{3} + 60 = 23 x + 2 x^{2}

x = 3, x = 4, x = - 5

Solution steps

x^{3} + 60 = 23 x + 2 x^{2}

Move

2 x^{2}

to the left side

x^{3} + 60 - 2 x^{2} = 23 x

Move

23 x

to the left side

x^{3} + 60 - 2 x^{2} - 23 x = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{3} - 2 x^{2} - 23 x + 60 = 0

Factor

x^{3} - 2 x^{2} - 23 x + 60 : (x - 3) (x - 4) (x + 5)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

x - 4 = 0 : x = 4

Solve

x + 5 = 0 : x = - 5

The solutions are

x = 3, x = 4, x = - 5

1 = m^{3}

x^{4} - 8 x^{2} = 16

\frac{r ^{3}}{9} = 3

x^{4} = - 2 + 23 i

5 - x^{3} = x^{2} + 6

What are the solutions to the equation x^3+60=23x+2x^2 ?
The solutions to the equation x^3+60=23x+2x^2 are x=3,x=4,x=-5
Find the zeros of x^3+60=23x+2x^2
The zeros of x^3+60=23x+2x^2 are x=3,x=4,x=-5