Find the Roots (Zeros) 5x^3+7x^2-11x+3

$5 x^{3} + 7 x^{2} - 11 x + 3$

Step 1

Set $5 x^{3} + 7 x^{2} - 11 x + 3$ equal to $0$ .

$5 x^{3} + 7 x^{2} - 11 x + 3 = 0$

Step 2

Solve for $x$ .

Factor $5 x^{3} + 7 x^{2} - 11 x + 3$ 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.

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

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

$\pm 1, \pm 0.2, \pm 3, \pm 0.6$

Substitute $0.6$ and simplify the expression. In this case, the expression is equal to $0$ so $0.6$ is a root of the polynomial.

Substitute $0.6$ into the polynomial.

$5 \cdot {0.6}^{3} + 7 \cdot {0.6}^{2} - 11 \cdot 0.6 + 3$

Raise $0.6$ to the power of $3$ .

$5 \cdot 0.216 + 7 \cdot {0.6}^{2} - 11 \cdot 0.6 + 3$

Multiply $5$ by $0.216$ .

$1.08 + 7 \cdot {0.6}^{2} - 11 \cdot 0.6 + 3$

Raise $0.6$ to the power of $2$ .

$1.08 + 7 \cdot 0.36 - 11 \cdot 0.6 + 3$

Multiply $7$ by $0.36$ .

$1.08 + 2.52 - 11 \cdot 0.6 + 3$

Add $1.08$ and $2.52$ .

$3.6 - 11 \cdot 0.6 + 3$

Multiply $- 11$ by $0.6$ .

$3.6 - 6.6 + 3$

Subtract $6.6$ from $3.6$ .

$- 3 + 3$

Add $- 3$ and $3$ .

$0$

Since $0.6$ is a known root, divide the polynomial by $5 x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{5 x^{3} + 7 x^{2} - 11 x + 3}{5 x - 3}$

Divide $5 x^{3} + 7 x^{2} - 11 x + 3$ by $5 x - 3$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$5 x$

$3$

$5 x^{3}$

$7 x^{2}$

$11 x$

$3$

Divide the highest order term in the dividend $5 x^{3}$ by the highest order term in divisor $5 x$ .

					$x^{2}$
	$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$

Multiply the new quotient term by the divisor.

				$x^{2}$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			+	$5 x^{3}$	-	$3 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{3} - 3 x^{2}$

				$x^{2}$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$

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

				$x^{2}$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$

Divide the highest order term in the dividend $10 x^{2}$ by the highest order term in divisor $5 x$ .

				$x^{2}$	+	$2 x$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					+	$10 x^{2}$	-	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x^{2} - 6 x$

				$x^{2}$	+	$2 x$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$

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

				$x^{2}$	+	$2 x$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$
							-	$5 x$

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

				$x^{2}$	+	$2 x$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$
							-	$5 x$	+	$3$

Divide the highest order term in the dividend $- 5 x$ by the highest order term in divisor $5 x$ .

				$x^{2}$	+	$2 x$	-	$1$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$
							-	$5 x$	+	$3$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	-	$1$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$
							-	$5 x$	+	$3$
							-	$5 x$	+	$3$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x + 3$

				$x^{2}$	+	$2 x$	-	$1$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$
							-	$5 x$	+	$3$
							+	$5 x$	-	$3$

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

				$x^{2}$	+	$2 x$	-	$1$
$5 x$	-	$3$		$5 x^{3}$	+	$7 x^{2}$	-	$11 x$	+	$3$
			-	$5 x^{3}$	+	$3 x^{2}$
					+	$10 x^{2}$	-	$11 x$
					-	$10 x^{2}$	+	$6 x$
							-	$5 x$	+	$3$
							+	$5 x$	-	$3$
										$0$

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

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

Write $5 x^{3} + 7 x^{2} - 11 x + 3$ as a set of factors.

$(5 x - 3) (x^{2} + 2 x - 1) = 0$

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

$5 x - 3 = 0$

$x^{2} + 2 x - 1 = 0$

Set $5 x - 3$ equal to $0$ and solve for $x$ .

Set $5 x - 3$ equal to $0$ .

$5 x - 3 = 0$

Solve $5 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$5 x = 3$

Divide each term in $5 x = 3$ by $5$ and simplify.

Divide each term in $5 x = 3$ by $5$ .

$\frac{5 x}{5} = \frac{3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{3}{5}$

Divide $x$ by $1$ .

$x = \frac{3}{5}$

Set $x^{2} + 2 x - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 2 x - 1$ equal to $0$ .

$x^{2} + 2 x - 1 = 0$

Solve $x^{2} + 2 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

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

$x = - 1 \pm \sqrt{2}$

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(5 x - 3) (x^{2} + 2 x - 1) = 0$ true.

$x = \frac{3}{5}, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3}{5}, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Decimal Form:

$x = 0.6, 0.41421356 \dots, - 2.41421356 \dots$

Step 4

Find the Roots (Zeros) 5x^3+7x^2-11x+3

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