Find the Roots/Zeros Using the Rational Roots Test x^2+7x+10

$x^{2} + 7 x + 10$

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.

$p = \pm 1, \pm 2, \pm 5, \pm 10$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 5, \pm 10$

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.

${(- 2)}^{2} + 7 (- 2) + 10$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 2$ is a root of the polynomial.

Simplify each term.

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

$4 + 7 (- 2) + 10$

Multiply $7$ by $- 2$ .

$4 - 14 + 10$

Simplify by adding and subtracting.

Subtract $14$ from $4$ .

$- 10 + 10$

Add $- 10$ and $10$ .

$0$

Step 5

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.

$\frac{x^{2} + 7 x + 10}{x + 2}$

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.

$- 2$	$1$	$7$	$10$
	\xa0		\xa0

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

$- 2$	$1$	$7$	$10$
	\xa0		\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 2)$ and place the result of $(- 2)$ under the next term in the dividend $(7)$ .

$- 2$	$1$	$7$	$10$
	\xa0	$- 2$	\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.

$- 2$	$1$	$7$	$10$
	\xa0	$- 2$	\xa0
	$1$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 2)$ and place the result of $(- 10)$ under the next term in the dividend $(10)$ .

$- 2$	$1$	$7$	$10$
	\xa0	$- 2$	$- 10$
	$1$	$5$

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

$- 2$	$1$	$7$	$10$
	\xa0	$- 2$	$- 10$
	$1$	$5$	$0$

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

$(1) x + 5$

Simplify the quotient polynomial.

$x + 5$

Step 7

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

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

$(x + 2) (x + 5)$

Step 9

These are the roots (zeros) of the polynomial $x^{2} + 7 x + 10$ .

$x = - 2, - 5$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^2+7x+10

Related Questions

Support LCMGCF.com by disabling your adblocker.