Find the Roots (Zeros) x^4+5x^3+8x^2+20x+16=0

$x^{4} + 5 x^{3} + 8 x^{2} + 20 x + 16 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$5 x^{3} + 20 x + x^{4} + 8 x^{2} + 16 = 0$

Factor $5 x$ out of $5 x^{3} + 20 x$ .

Factor $5 x$ out of $5 x^{3}$ .

$5 x (x^{2}) + 20 x + x^{4} + 8 x^{2} + 16 = 0$

Factor $5 x$ out of $20 x$ .

$5 x (x^{2}) + 5 x (4) + x^{4} + 8 x^{2} + 16 = 0$

Factor $5 x$ out of $5 x (x^{2}) + 5 x (4)$ .

$5 x (x^{2} + 4) + x^{4} + 8 x^{2} + 16 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$5 x (x^{2} + 4) + {(x^{2})}^{2} + 8 x^{2} + 16 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$5 x (x^{2} + 4) + u^{2} + 8 u + 16 = 0$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$5 x (x^{2} + 4) + u^{2} + 8 u + 4^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 u = 2 \cdot u \cdot 4$

Rewrite the polynomial.

$5 x (x^{2} + 4) + u^{2} + 2 \cdot u \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 4$ .

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

Replace all occurrences of $u$ with $x^{2}$ .

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

Factor $x^{2} + 4$ out of $5 x (x^{2} + 4) + {(x^{2} + 4)}^{2}$ .

Factor $x^{2} + 4$ out of $5 x (x^{2} + 4)$ .

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

Factor $x^{2} + 4$ out of ${(x^{2} + 4)}^{2}$ .

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

Factor $x^{2} + 4$ out of $(x^{2} + 4) (5 x) + (x^{2} + 4) (x^{2} + 4)$ .

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

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

Factor $5 u + u^{2} + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

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

Factor.

Replace all occurrences of $u$ with $x$ .

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

Remove unnecessary parentheses.

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

Step 2

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

$x^{2} + 4 = 0$

$x + 1 = 0$

$x + 4 = 0$

Step 3

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4$ equal to $0$ .

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{4}$

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

Step 4

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 5

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

The final solution is all the values that make $(x^{2} + 4) (x + 1) (x + 4) = 0$ true.

$x = 2 i, - 2 i, - 1, - 4$

Step 7

Find the Roots (Zeros) x^4+5x^3+8x^2+20x+16=0

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