Solve for x x^3-5x^2+x-5=x^2+1

$x^{3} - 5 x^{2} + x - 5 = x^{2} + 1$

Step 1

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$x^{3} - 5 x^{2} + x - 5 - x^{2} = 1$

Subtract $x^{2}$ from $- 5 x^{2}$ .

$x^{3} - 6 x^{2} + x - 5 = 1$

Step 2

Subtract $1$ from both sides of the equation.

$x^{3} - 6 x^{2} + x - 5 - 1 = 0$

Step 3

Subtract $1$ from $- 5$ .

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

Step 4

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 6 x^{2}) + x - 6 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 6) + 1 (x - 6) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 6$ .

$(x - 6) (x^{2} + 1) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 6 = 0$

$x^{2} + 1 = 0$

Step 6

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 7

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 8

The final solution is all the values that make $(x - 6) (x^{2} + 1) = 0$ true.

$x = 6, i, - i$

Solve for x x^3-5x^2+x-5=x^2+1

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