What are the solutions to the equation x^3-2x^2+8=3x^2+10x-42 ?

The solutions to the equation x^3-2x^2+8=3x^2+10x-42 are x=5,x=-sqrt(10),x=sqrt(10)

Find the zeros of x^3-2x^2+8=3x^2+10x-42

The zeros of x^3-2x^2+8=3x^2+10x-42 are x=5,x=-sqrt(10),x=sqrt(10)

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x^{3} - 2 x^{2} + 8 = 3 x^{2} + 10 x - 42

x = 5, x = - 10, x = 10

Decimal Notation

x = 5, x = - 3.16227 \dots, x = 3.16227 \dots

Solution steps

x^{3} - 2 x^{2} + 8 = 3 x^{2} + 10 x - 42

Move

42

to the left side

x^{3} - 2 x^{2} + 50 = 3 x^{2} + 10 x

Move

10 x

to the left side

x^{3} - 2 x^{2} + 50 - 10 x = 3 x^{2}

Move

3 x^{2}

to the left side

x^{3} - 5 x^{2} - 10 x + 50 = 0

Factor

x^{3} - 5 x^{2} - 10 x + 50 : (x - 5) (x + 10) (x - 10)

(x - 5) (x + 10) (x - 10) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 5 = 0 or x + 10 = 0 or x - 10 = 0

Solve

x - 5 = 0 : x = 5

Solve

x + 10 = 0 : x = - 10

Solve

x - 10 = 0 : x = 10

The solutions are

x = 5, x = - 10, x = 10

20 x^{4} (1 - x)^{3} = 0

c^{3} - c^{2} + c - 2 = 4

(2 x^{2} + 1) x^{3} = 12

(z^{2} + 6)^{4} = 0

x^{3} - 4 x^{2} + 7 x - 4 = 0

What are the solutions to the equation x^3-2x^2+8=3x^2+10x-42 ?
The solutions to the equation x^3-2x^2+8=3x^2+10x-42 are x=5,x=-sqrt(10),x=sqrt(10)
Find the zeros of x^3-2x^2+8=3x^2+10x-42
The zeros of x^3-2x^2+8=3x^2+10x-42 are x=5,x=-sqrt(10),x=sqrt(10)