Find the Roots (Zeros) f(x)=5x^3+5x^2-170x+280

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

Step 1

Set $5 x^{3} + 5 x^{2} - 170 x + 280$ equal to $0$ .

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

Step 2

Solve for $x$ .

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.

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

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

$x - 2 = 0$

$x - 4 = 0$

$x + 7 = 0$

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.

$x = 2$

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.

$x = 4$

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.

$x = - 7$

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

$x = 2, 4, - 7$

Step 3

Find the Roots (Zeros) f(x)=5x^3+5x^2-170x+280

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