What are the solutions to the equation x^3-x=x^2+5x-6 ?

The solutions to the equation x^3-x=x^2+5x-6 are x=1,x=-sqrt(6),x=sqrt(6)

Find the zeros of x^3-x=x^2+5x-6

The zeros of x^3-x=x^2+5x-6 are x=1,x=-sqrt(6),x=sqrt(6)

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

x = 1, x = - 6, x = 6

Decimal Notation

x = 1, x = - 2.44948 \dots, x = 2.44948 \dots

Solution steps

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

Move

6

to the left side

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

Move

5 x

to the left side

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

Move

x^{2}

to the left side

x^{3} - 6 x + 6 - x^{2} = 0

Write in the standard form

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

x^{3} - x^{2} - 6 x + 6 = 0

Factor

x^{3} - x^{2} - 6 x + 6 : (x - 1) (x + 6) (x - 6)

(x - 1) (x + 6) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 6 = 0 or x - 6 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 6 = 0 : x = - 6

Solve

x - 6 = 0 : x = 6

The solutions are

x = 1, x = - 6, x = 6

x^{4} - 81

x^{6} - 8 x^{5} + 12 x^{4} = 0

x^{3} - 4 x^{2} - 3 x + 6 = 0

2 x^{4} - x^{3} - 14 x^{2} + 19 x - 6 = 0

- 3 x^{4} + 15 x^{2} - 12 = 0

What are the solutions to the equation x^3-x=x^2+5x-6 ?
The solutions to the equation x^3-x=x^2+5x-6 are x=1,x=-sqrt(6),x=sqrt(6)
Find the zeros of x^3-x=x^2+5x-6
The zeros of x^3-x=x^2+5x-6 are x=1,x=-sqrt(6),x=sqrt(6)