Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+7x^2-9x-15

$f (x) = x^{3} + 7 x^{2} - 9 x - 15$

Step 1

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 5, \pm 15$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

${(- 1)}^{3} + 7 {(- 1)}^{2} - 9 \cdot - 1 - 15$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

Raise $- 1$ to the power of $3$ .

$- 1 + 7 {(- 1)}^{2} - 9 \cdot - 1 - 15$

Raise $- 1$ to the power of $2$ .

$- 1 + 7 \cdot 1 - 9 \cdot - 1 - 15$

Multiply $7$ by $1$ .

$- 1 + 7 - 9 \cdot - 1 - 15$

Multiply $- 9$ by $- 1$ .

$- 1 + 7 + 9 - 15$

Simplify by adding and subtracting.

Add $- 1$ and $7$ .

$6 + 9 - 15$

Add $6$ and $9$ .

$15 - 15$

Subtract $15$ from $15$ .

$0$

Step 5

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.

$\frac{x^{3} + 7 x^{2} - 9 x - 15}{x + 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0			\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(7)$ .

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0	$- 1$		\xa0
	$1$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0	$- 1$		\xa0
	$1$	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 1)$ and place the result of $(- 6)$ under the next term in the dividend $(- 9)$ .

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0	$- 1$	$- 6$	\xa0
	$1$	$6$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0	$- 1$	$- 6$	\xa0
	$1$	$6$	$- 15$

Multiply the newest entry in the result $(- 15)$ by the divisor $(- 1)$ and place the result of $(15)$ under the next term in the dividend $(- 15)$ .

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0	$- 1$	$- 6$	$15$
	$1$	$6$	$- 15$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$1$	$7$	$- 9$	$- 15$
	\xa0	$- 1$	$- 6$	$15$
	$1$	$6$	$- 15$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$1 x^{2} + 6 x - 15$

Simplify the quotient polynomial.

$x^{2} + 6 x - 15$

Step 7

Solve the equation to find any remaining roots.

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 6$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot - 15)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 15$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$x = \frac{- 6 \pm \sqrt{36 + 60}}{2 \cdot 1}$

Add $36$ and $60$ .

$x = \frac{- 6 \pm \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$x = \frac{- 6 \pm \sqrt{16 (6)}}{2 \cdot 1}$

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

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

$x = - 3 \pm 2 \sqrt{6}$

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

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 15$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$x = \frac{- 6 \pm \sqrt{36 + 60}}{2 \cdot 1}$

Add $36$ and $60$ .

$x = \frac{- 6 \pm \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$x = \frac{- 6 \pm \sqrt{16 (6)}}{2 \cdot 1}$

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

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

$x = - 3 \pm 2 \sqrt{6}$

Change the $\pm$ to $+$ .

$x = - 3 + 2 \sqrt{6}$

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

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 15$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$x = \frac{- 6 \pm \sqrt{36 + 60}}{2 \cdot 1}$

Add $36$ and $60$ .

$x = \frac{- 6 \pm \sqrt{96}}{2 \cdot 1}$

Rewrite $96$ as $4^{2} \cdot 6$ .

Factor $16$ out of $96$ .

$x = \frac{- 6 \pm \sqrt{16 (6)}}{2 \cdot 1}$

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

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

$x = - 3 \pm 2 \sqrt{6}$

Change the $\pm$ to $-$ .

$x = - 3 - 2 \sqrt{6}$

The final answer is the combination of both solutions.

$x = - 3 + 2 \sqrt{6}, - 3 - 2 \sqrt{6}$

Step 8

The polynomial can be written as a set of linear factors.

$(x + 1) (x + 3 - 2 \sqrt{6}) (x + 3 + 2 \sqrt{6})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} + 7 x^{2} - 9 x - 15$ .

$x = - 1, - 3 + 2 \sqrt{6}, - 3 - 2 \sqrt{6}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = - 1, - 3 + 2 \sqrt{6}, - 3 - 2 \sqrt{6}$

Decimal Form:

$x = - 1, 1.89897948 \dots, - 7.89897948 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+7x^2-9x-15

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