Find the Roots/Zeros Using the Rational Roots Test x^3-11x^2+21x+49

$x^{3} - 11 x^{2} + 21 x + 49$

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.

$p = \pm 1, \pm 7, \pm 49$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 7, \pm 49$

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.

${(7)}^{3} - 11 {(7)}^{2} + 21 (7) + 49$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 7$ is a root of the polynomial.

Simplify each term.

Raise $7$ to the power of $3$ .

$343 - 11 {(7)}^{2} + 21 (7) + 49$

Raise $7$ to the power of $2$ .

$343 - 11 \cdot 49 + 21 (7) + 49$

Multiply $- 11$ by $49$ .

$343 - 539 + 21 (7) + 49$

Multiply $21$ by $7$ .

$343 - 539 + 147 + 49$

Simplify by adding and subtracting.

Subtract $539$ from $343$ .

$- 196 + 147 + 49$

Add $- 196$ and $147$ .

$- 49 + 49$

Add $- 49$ and $49$ .

$0$

Step 5

Since $7$ is a known root, divide the polynomial by $x - 7$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 11 x^{2} + 21 x + 49}{x - 7}$

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.

$7$	$1$	$- 11$	$21$	$49$
	\xa0			\xa0

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

$7$	$1$	$- 11$	$21$	$49$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(7)$ and place the result of $(7)$ under the next term in the dividend $(- 11)$ .

$7$	$1$	$- 11$	$21$	$49$
	\xa0	$7$		\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.

$7$	$1$	$- 11$	$21$	$49$
	\xa0	$7$		\xa0
	$1$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(7)$ and place the result of $(- 28)$ under the next term in the dividend $(21)$ .

$7$	$1$	$- 11$	$21$	$49$
	\xa0	$7$	$- 28$	\xa0
	$1$	$- 4$

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

$7$	$1$	$- 11$	$21$	$49$
	\xa0	$7$	$- 28$	\xa0
	$1$	$- 4$	$- 7$

Multiply the newest entry in the result $(- 7)$ by the divisor $(7)$ and place the result of $(- 49)$ under the next term in the dividend $(49)$ .

$7$	$1$	$- 11$	$21$	$49$
	\xa0	$7$	$- 28$	$- 49$
	$1$	$- 4$	$- 7$

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

$7$	$1$	$- 11$	$21$	$49$
	\xa0	$7$	$- 28$	$- 49$
	$1$	$- 4$	$- 7$	$0$

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

$1 x^{2} + - 4 x - 7$

Simplify the quotient polynomial.

$x^{2} - 4 x - 7$

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 $a = 1$ , $b = - 4$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot - 7)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 7$ .

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 7}}{2 \cdot 1}$

Multiply $- 4$ by $- 7$ .

$x = \frac{4 \pm \sqrt{16 + 28}}{2 \cdot 1}$

Add $16$ and $28$ .

$x = \frac{4 \pm \sqrt{44}}{2 \cdot 1}$

Rewrite $44$ as $2^{2} \cdot 11$ .

Factor $4$ out of $44$ .

$x = \frac{4 \pm \sqrt{4 (11)}}{2 \cdot 1}$

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 11}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{11}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{11}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{11}}{2}$ .

$x = 2 \pm \sqrt{11}$

The final answer is the combination of both solutions.

$x = 2 + \sqrt{11}, 2 - \sqrt{11}$

Step 8

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

$(x - 7) (x - 2 - \sqrt{11}) (x - 2 + \sqrt{11})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - 11 x^{2} + 21 x + 49$ .

$x = 7, 2 + \sqrt{11}, 2 - \sqrt{11}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = 7, 2 + \sqrt{11}, 2 - \sqrt{11}$

Decimal Form:

$x = 7, 5.31662479 \dots, - 1.31662479 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test x^3-11x^2+21x+49

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