Find the Roots/Zeros Using the Rational Roots Test 5x^3+5x^2-170x+280

$5 x^{3} + 5 x^{2} - 170 x + 280$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 7, \pm 8, \pm 10, \pm 14, \pm 20, \pm 28, \pm 35, \pm 40, \pm 56, \pm 70, \pm 140, \pm 280$

$q = \pm 1, \pm 5$

Step 2

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

$\pm 1, \pm \frac{1}{5}, \pm 2, \pm \frac{2}{5}, \pm 4, \pm \frac{4}{5}, \pm 5, \pm 7, \pm \frac{7}{5}, \pm 8, \pm \frac{8}{5}, \pm 10, \pm 14, \pm \frac{14}{5}, \pm 20, \pm 28, \pm \frac{28}{5}, \pm 35, \pm 40, \pm 56, \pm \frac{56}{5}, \pm 70, \pm 140, \pm 280$

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.

$5 {(2)}^{3} + 5 {(2)}^{2} - 170 \cdot 2 + 280$

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 $3$ .

$5 \cdot 8 + 5 {(2)}^{2} - 170 \cdot 2 + 280$

Multiply $5$ by $8$ .

$40 + 5 {(2)}^{2} - 170 \cdot 2 + 280$

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

$40 + 5 \cdot 4 - 170 \cdot 2 + 280$

Multiply $5$ by $4$ .

$40 + 20 - 170 \cdot 2 + 280$

Multiply $- 170$ by $2$ .

$40 + 20 - 340 + 280$

Simplify by adding and subtracting.

Add $40$ and $20$ .

$60 - 340 + 280$

Subtract $340$ from $60$ .

$- 280 + 280$

Add $- 280$ and $280$ .

$0$

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.

$\frac{5 x^{3} + 5 x^{2} - 170 x + 280}{x - 2}$

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.

$2$	$5$	$5$	$- 170$	$280$
	\xa0			\xa0

The first number in the dividend $(5)$ is put into the first position of the result area (below the horizontal line).

$2$	$5$	$5$	$- 170$	$280$
	\xa0			\xa0
	$5$

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 $(5)$ .

$2$	$5$	$5$	$- 170$	$280$
	\xa0	$10$		\xa0
	$5$

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

$2$	$5$	$5$	$- 170$	$280$
	\xa0	$10$		\xa0
	$5$	$15$

Multiply the newest entry in the result $(15)$ by the divisor $(2)$ and place the result of $(30)$ under the next term in the dividend $(- 170)$ .

$2$	$5$	$5$	$- 170$	$280$
	\xa0	$10$	$30$	\xa0
	$5$	$15$

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

$2$	$5$	$5$	$- 170$	$280$
	\xa0	$10$	$30$	\xa0
	$5$	$15$	$- 140$

Multiply the newest entry in the result $(- 140)$ by the divisor $(2)$ and place the result of $(- 280)$ under the next term in the dividend $(280)$ .

$2$	$5$	$5$	$- 170$	$280$
	\xa0	$10$	$30$	$- 280$
	$5$	$15$	$- 140$

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

$2$	$5$	$5$	$- 170$	$280$
	\xa0	$10$	$30$	$- 280$
	$5$	$15$	$- 140$	$0$

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

$5 x^{2} + 15 x - 140$

Step 7

Factor $5$ out of $5 x^{2} + 15 x - 140$ .

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

$(x - 2) (5 (x^{2}) + 15 x - 140)$

Factor $5$ out of $15 x$ .

$(x - 2) (5 (x^{2}) + 5 (3 x) - 140)$

Factor $5$ out of $- 140$ .

$(x - 2) (5 x^{2} + 5 (3 x) + 5 \cdot - 28)$

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

$(x - 2) (5 (x^{2} + 3 x) + 5 \cdot - 28)$

Factor $5$ out of $5 (x^{2} + 3 x) + 5 \cdot - 28$ .

$(x - 2) (5 (x^{2} + 3 x - 28))$

Step 8

Factor.

Factor $x^{2} + 3 x - 28$ 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 $- 28$ and whose sum is $3$ .

$- 4, 7$

Write the factored form using these integers.

$(x - 2) (5 ((x - 4) (x + 7)))$

Remove unnecessary parentheses.

$(x - 2) (5 (x - 4) (x + 7))$

Step 9

Factor the left side of the equation.

Factor $5$ out of $5 x^{3} + 5 x^{2} - 170 x + 280$ .

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

$5 (x^{3}) + 5 x^{2} - 170 x + 280 = 0$

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

$5 (x^{3}) + 5 (x^{2}) - 170 x + 280 = 0$

Factor $5$ out of $- 170 x$ .

$5 (x^{3}) + 5 (x^{2}) + 5 (- 34 x) + 280 = 0$

Factor $5$ out of $280$ .

$5 (x^{3}) + 5 x^{2} + 5 (- 34 x) + 5 \cdot 56 = 0$

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

$5 (x^{3} + x^{2}) + 5 (- 34 x) + 5 \cdot 56 = 0$

Factor $5$ out of $5 (x^{3} + x^{2}) + 5 (- 34 x)$ .

$5 (x^{3} + x^{2} - 34 x) + 5 \cdot 56 = 0$

Factor $5$ out of $5 (x^{3} + x^{2} - 34 x) + 5 \cdot 56$ .

$5 (x^{3} + x^{2} - 34 x + 56) = 0$

Factor $x^{3} + x^{2} - 34 x + 56$ 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.

$p = \pm 1, \pm 56, \pm 2, \pm 28, \pm 4, \pm 14, \pm 7, \pm 8 q = \pm 1$

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

$\pm 1, \pm 56, \pm 2, \pm 28, \pm 4, \pm 14, \pm 7, \pm 8$

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.

$2^{3} + 2^{2} - 34 \cdot 2 + 56$

Raise $2$ to the power of $3$ .

$8 + 2^{2} - 34 \cdot 2 + 56$

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

$8 + 4 - 34 \cdot 2 + 56$

Add $8$ and $4$ .

$12 - 34 \cdot 2 + 56$

Multiply $- 34$ by $2$ .

$12 - 68 + 56$

Subtract $68$ from $12$ .

$- 56 + 56$

Add $- 56$ and $56$ .

$0$

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.

$\frac{x^{3} + x^{2} - 34 x + 56}{x - 2}$

Divide $x^{3} + x^{2} - 34 x + 56$ 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 $0$ .

$x$

$2$

$x^{3}$

$x^{2}$

$34 x$

$56$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					+	$3 x^{2}$	-	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $3 x^{2} - 6 x$

				$x^{2}$	+	$3 x$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$
							-	$28 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$
							-	$28 x$	+	$56$

Divide the highest order term in the dividend $- 28 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$
							-	$28 x$	+	$56$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$
							-	$28 x$	+	$56$
							-	$28 x$	+	$56$

The expression needs to be subtracted from the dividend, so change all the signs in $- 28 x + 56$

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$
							-	$28 x$	+	$56$
							+	$28 x$	-	$56$

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

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$2$		$x^{3}$	+	$x^{2}$	-	$34 x$	+	$56$
			-	$x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$34 x$
					-	$3 x^{2}$	+	$6 x$
							-	$28 x$	+	$56$
							+	$28 x$	-	$56$
										$0$

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

$x^{2} + 3 x - 28$

Write $x^{3} + x^{2} - 34 x + 56$ as a set of factors.

$5 ((x - 2) (x^{2} + 3 x - 28)) = 0$

Factor.

Factor $x^{2} + 3 x - 28$ 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 $- 28$ and whose sum is $3$ .

$- 4, 7$

Write the factored form using these integers.

$5 ((x - 2) ((x - 4) (x + 7))) = 0$

Remove unnecessary parentheses.

$5 ((x - 2) (x - 4) (x + 7)) = 0$

Remove unnecessary parentheses.

$5 (x - 2) (x - 4) (x + 7) = 0$

Step 10

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

$x - 2 = 0$

$x - 4 = 0$

$x + 7 = 0$

Step 11

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 12

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 13

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 14

The final solution is all the values that make $5 (x - 2) (x - 4) (x + 7) = 0$ true.

$x = 2, 4, - 7$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 5x^3+5x^2-170x+280

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