What are the solutions to the equation x^3-4x^2+x-4=x^2+1 ?

The solutions to the equation x^3-4x^2+x-4=x^2+1 are x=5,x=i,x=-i

Find the zeros of x^3-4x^2+x-4=x^2+1

The zeros of x^3-4x^2+x-4=x^2+1 are x=5,x=i,x=-i

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x^{3} - 4 x^{2} + x - 4 = x^{2} + 1

x = 5, x = i, x = - i

Solution steps

x^{3} - 4 x^{2} + x - 4 = x^{2} + 1

Move

1

to the left side

x^{3} - 4 x^{2} + x - 5 = x^{2}

Move

x^{2}

to the left side

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

Factor

x^{3} - 5 x^{2} + x - 5 : (x - 5) (x^{2} + 1)

(x - 5) (x^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 5 = 0 or x^{2} + 1 = 0

Solve

x - 5 = 0 : x = 5

Solve

x^{2} + 1 = 0 : x = i, x = - i

The solutions are

x = 5, x = i, x = - i

4 x^{3} + 24 x^{2} + 36 x = 0

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

x^{3} - 6 x^{2} + 9 x + 2 = 0

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

2 x^{3} - x = 0

What are the solutions to the equation x^3-4x^2+x-4=x^2+1 ?
The solutions to the equation x^3-4x^2+x-4=x^2+1 are x=5,x=i,x=-i
Find the zeros of x^3-4x^2+x-4=x^2+1
The zeros of x^3-4x^2+x-4=x^2+1 are x=5,x=i,x=-i