Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+4x^3+5x^2-20x-50

$f (x) = x^{4} + 4 x^{3} + 5 x^{2} - 20 x - 50$

Step 1

Factor the left side of the equation.

Regroup terms.

$4 x^{3} - 20 x + x^{4} + 5 x^{2} - 50 = 0$

Factor $4 x$ out of $4 x^{3} - 20 x$ .

Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 20 x + x^{4} + 5 x^{2} - 50 = 0$

Factor $4 x$ out of $- 20 x$ .

$4 x (x^{2}) + 4 x (- 5) + x^{4} + 5 x^{2} - 50 = 0$

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

$4 x (x^{2} - 5) + x^{4} + 5 x^{2} - 50 = 0$

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

$4 x (x^{2} - 5) + {(x^{2})}^{2} + 5 x^{2} - 50 = 0$

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

$4 x (x^{2} - 5) + u^{2} + 5 u - 50 = 0$

Factor $u^{2} + 5 u - 50$ 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 $- 50$ and whose sum is $5$ .

$- 5, 10$

Write the factored form using these integers.

$4 x (x^{2} - 5) + (u - 5) (u + 10) = 0$

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

$4 x (x^{2} - 5) + (x^{2} - 5) (x^{2} + 10) = 0$

Factor $x^{2} - 5$ out of $4 x (x^{2} - 5) + (x^{2} - 5) (x^{2} + 10)$ .

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

$(x^{2} - 5) (4 x) + (x^{2} - 5) (x^{2} + 10) = 0$

Factor $x^{2} - 5$ out of $(x^{2} - 5) (4 x) + (x^{2} - 5) (x^{2} + 10)$ .

$(x^{2} - 5) (4 x + x^{2} + 10) = 0$

Reorder terms.

$(x^{2} - 5) (x^{2} + 4 x + 10) = 0$

Step 2

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

$x^{2} - 5 = 0$

$x^{2} + 4 x + 10 = 0$

Step 3

Set $x^{2} - 5$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5$ equal to $0$ .

$x^{2} - 5 = 0$

Solve $x^{2} - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$x^{2} = 5$

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

$x = \pm \sqrt{5}$

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.

$x = \sqrt{5}$

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

$x = - \sqrt{5}$

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

$x = \sqrt{5}, - \sqrt{5}$

Step 4

Set $x^{2} + 4 x + 10$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4 x + 10$ equal to $0$ .

$x^{2} + 4 x + 10 = 0$

Solve $x^{2} + 4 x + 10 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 4$ , and $c = 10$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 10$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 10}}{2 \cdot 1}$

Multiply $- 4$ by $10$ .

$x = \frac{- 4 \pm \sqrt{16 - 40}}{2 \cdot 1}$

Subtract $40$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 24}}{2 \cdot 1}$

Rewrite $- 24$ as $- 1 (24)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 24}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (6)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{6})}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 2 i \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 i \sqrt{6}}{2}$

Simplify $\frac{- 4 \pm 2 i \sqrt{6}}{2}$ .

$x = - 2 \pm i \sqrt{6}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 10$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 10}}{2 \cdot 1}$

Multiply $- 4$ by $10$ .

$x = \frac{- 4 \pm \sqrt{16 - 40}}{2 \cdot 1}$

Subtract $40$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 24}}{2 \cdot 1}$

Rewrite $- 24$ as $- 1 (24)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 24}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (6)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{6})}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 2 i \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 i \sqrt{6}}{2}$

Simplify $\frac{- 4 \pm 2 i \sqrt{6}}{2}$ .

$x = - 2 \pm i \sqrt{6}$

Change the $\pm$ to $+$ .

$x = - 2 + i \sqrt{6}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 10$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 10}}{2 \cdot 1}$

Multiply $- 4$ by $10$ .

$x = \frac{- 4 \pm \sqrt{16 - 40}}{2 \cdot 1}$

Subtract $40$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 24}}{2 \cdot 1}$

Rewrite $- 24$ as $- 1 (24)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 24}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (6)}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{6})}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 2 i \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 i \sqrt{6}}{2}$

Simplify $\frac{- 4 \pm 2 i \sqrt{6}}{2}$ .

$x = - 2 \pm i \sqrt{6}$

Change the $\pm$ to $-$ .

$x = - 2 - i \sqrt{6}$

The final answer is the combination of both solutions.

$x = - 2 + i \sqrt{6}, - 2 - i \sqrt{6}$

Step 5

The final solution is all the values that make $(x^{2} - 5) (x^{2} + 4 x + 10) = 0$ true.

$x = \sqrt{5}, - \sqrt{5}, - 2 + i \sqrt{6}, - 2 - i \sqrt{6}$

Step 6

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+4x^3+5x^2-20x-50

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