Find the Roots (Zeros) f(x)=x^4+5x^3+10x^2+20x+24

$f (x) = x^{4} + 5 x^{3} + 10 x^{2} + 20 x + 24$

Step 1

Set $x^{4} + 5 x^{3} + 10 x^{2} + 20 x + 24$ equal to $0$ .

$x^{4} + 5 x^{3} + 10 x^{2} + 20 x + 24 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$5 x^{3} + 10 x^{2} + x^{4} + 20 x + 24 = 0$

Factor $5 x^{2}$ out of $5 x^{3} + 10 x^{2}$ .

Factor $5 x^{2}$ out of $5 x^{3}$ .

$5 x^{2} (x) + 10 x^{2} + x^{4} + 20 x + 24 = 0$

Factor $5 x^{2}$ out of $10 x^{2}$ .

$5 x^{2} (x) + 5 x^{2} (2) + x^{4} + 20 x + 24 = 0$

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

$5 x^{2} (x + 2) + x^{4} + 20 x + 24 = 0$

Factor $x^{4} + 20 x + 24$ 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.

$p = \pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6 q = \pm 1$

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

$\pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6$

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

${(- 2)}^{4} + 20 \cdot - 2 + 24$

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

$16 + 20 \cdot - 2 + 24$

Multiply $20$ by $- 2$ .

$16 - 40 + 24$

Subtract $40$ from $16$ .

$- 24 + 24$

Add $- 24$ and $24$ .

$0$

Since $- 2$ is a known root, divide the polynomial by $x + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} + 20 x + 24}{x + 2}$

Divide $x^{4} + 20 x + 24$ by $x + 2$ .

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

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

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

$x^{3}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{4} + 2 x^{3}$

$x^{3}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

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

$x^{3}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

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

$x^{3}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

Divide the highest order term in the dividend $- 2 x^{3}$ by the highest order term in divisor $x$ .

$x^{3}$

$2 x^{2}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

Multiply the new quotient term by the divisor.

$x^{3}$

$2 x^{2}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{3} - 4 x^{2}$

$x^{3}$

$2 x^{2}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

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

$x^{3}$

$2 x^{2}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

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

$x^{3}$

$2 x^{2}$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

Divide the highest order term in the dividend $4 x^{2}$ by the highest order term in divisor $x$ .

$x^{3}$

$2 x^{2}$

$4 x$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$2 x^{2}$

$4 x$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{2} + 8 x$

$x^{3}$

$2 x^{2}$

$4 x$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

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

$x^{3}$

$2 x^{2}$

$4 x$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

$12 x$

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

$x^{3}$

$2 x^{2}$

$4 x$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

$12 x$

$24$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

$x^{3}$

$2 x^{2}$

$4 x$

$12$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

$12 x$

$24$

Multiply the new quotient term by the divisor.

$x^{3}$

$2 x^{2}$

$4 x$

$12$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

$12 x$

$24$

$12 x$

$24$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 24$

$x^{3}$

$2 x^{2}$

$4 x$

$12$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

$12 x$

$24$

$12 x$

$24$

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

$x^{3}$

$2 x^{2}$

$4 x$

$12$

$x$

$2$

$x^{4}$

$0 x^{3}$

$0 x^{2}$

$20 x$

$24$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x^{3}$

$4 x^{2}$

$20 x$

$4 x^{2}$

$8 x$

$12 x$

$24$

$12 x$

$24$

$0$

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

$x^{3} - 2 x^{2} + 4 x + 12$

Write $x^{4} + 20 x + 24$ as a set of factors.

$5 x^{2} (x + 2) + (x + 2) (x^{3} - 2 x^{2} + 4 x + 12) = 0$

Factor $x + 2$ out of $5 x^{2} (x + 2) + (x + 2) (x^{3} - 2 x^{2} + 4 x + 12)$ .

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

$(x + 2) (5 x^{2}) + (x + 2) (x^{3} - 2 x^{2} + 4 x + 12) = 0$

Factor $x + 2$ out of $(x + 2) (5 x^{2}) + (x + 2) (x^{3} - 2 x^{2} + 4 x + 12)$ .

$(x + 2) (5 x^{2} + x^{3} - 2 x^{2} + 4 x + 12) = 0$

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

$(x + 2) (x^{3} + 3 x^{2} + 4 x + 12) = 0$

Factor.

Rewrite $x^{3} + 3 x^{2} + 4 x + 12$ in a factored form.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 2) ((x^{3} + 3 x^{2}) + 4 x + 12) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

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

Remove unnecessary parentheses.

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

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

$x + 2 = 0$

$x + 3 = 0$

$x^{2} + 4 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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.

$x = 2 i, - 2 i$

The final solution is all the values that make $(x + 2) (x + 3) (x^{2} + 4) = 0$ true.

$x = - 2, - 3, 2 i, - 2 i$

Step 3

Find the Roots (Zeros) f(x)=x^4+5x^3+10x^2+20x+24

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