Find All Complex Solutions x^4+6x^3+6x^2+6x+5=0

$x^{4} + 6 x^{3} + 6 x^{2} + 6 x + 5 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$6 x^{3} + 6 x^{2} + x^{4} + 6 x + 5 = 0$

Factor $6 x^{2}$ out of $6 x^{3} + 6 x^{2}$ .

Factor $6 x^{2}$ out of $6 x^{3}$ .

$6 x^{2} (x) + 6 x^{2} + x^{4} + 6 x + 5 = 0$

Factor $6 x^{2}$ out of $6 x^{2}$ .

$6 x^{2} (x) + 6 x^{2} (1) + x^{4} + 6 x + 5 = 0$

Factor $6 x^{2}$ out of $6 x^{2} (x) + 6 x^{2} (1)$ .

$6 x^{2} (x + 1) + x^{4} + 6 x + 5 = 0$

Factor $x^{4} + 6 x + 5$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 5 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 5$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

${(- 1)}^{4} + 6 \cdot - 1 + 5$

Raise $- 1$ to the power of $4$ .

$1 + 6 \cdot - 1 + 5$

Multiply $6$ by $- 1$ .

$1 - 6 + 5$

Subtract $6$ from $1$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} + 6 x + 5}{x + 1}$

Divide $x^{4} + 6 x + 5$ by $x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

Divide the highest order term in the dividend $x^{4}$ by the highest order term in divisor $x$ .

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{4} + x^{3}$

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

Pull the next terms from the original dividend down into the current dividend.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

Divide the highest order term in the dividend $- x^{3}$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{3} - x^{2}$

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

Pull the next terms from the original dividend down into the current dividend.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + x$

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

$5 x$

Pull the next terms from the original dividend down into the current dividend.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

$5 x$

$5$

Divide the highest order term in the dividend $5 x$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$x$

$5$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

$5 x$

$5$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$5$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

$5 x$

$5$

$5 x$

$5$

The expression needs to be subtracted from the dividend, so change all the signs in $5 x + 5$

$x^{3}$

$x^{2}$

$x$

$5$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

$5 x$

$5$

$5 x$

$5$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{3}$

$x^{2}$

$x$

$5$

$x$

$1$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$6 x$

$5$

$x^{4}$

$x^{3}$

$0 x^{2}$

$x^{3}$

$x^{2}$

$6 x$

$x^{2}$

$x$

$5 x$

$5$

$5 x$

$5$

$0$

Since the remander is $0$ , the final answer is the quotient.

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

Write $x^{4} + 6 x + 5$ as a set of factors.

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

Factor $x + 1$ out of $6 x^{2} (x + 1) + (x + 1) (x^{3} - x^{2} + x + 5)$ .

Factor $x + 1$ out of $6 x^{2} (x + 1)$ .

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

Factor $x + 1$ out of $(x + 1) (6 x^{2}) + (x + 1) (x^{3} - x^{2} + x + 5)$ .

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

Subtract $x^{2}$ from $6 x^{2}$ .

$(x + 1) (x^{3} + 5 x^{2} + x + 5) = 0$

Factor.

Rewrite $x^{3} + 5 x^{2} + x + 5$ in a factored form.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((x^{3} + 5 x^{2}) + x + 5) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $x + 5$ .

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

Remove unnecessary parentheses.

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

Step 2

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

$x + 1 = 0$

$x + 5 = 0$

$x^{2} + 1 = 0$

Step 3

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 5

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 specified 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 6

The final solution is all the values that make $(x + 1) (x + 5) (x^{2} + 1) = 0$ true.

$x = - 1, - 5, i, - i$

Find All Complex Solutions x^4+6x^3+6x^2+6x+5=0

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