Find the Roots (Zeros) 2x^3-5x^2+2x+1

$2 x^{3} - 5 x^{2} + 2 x + 1$

Step 1

Set $2 x^{3} - 5 x^{2} + 2 x + 1$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor $2 x^{3} - 5 x^{2} + 2 x + 1$ 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.

$p = \pm 1 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$2 \cdot 1^{3} - 5 \cdot 1^{2} + 2 \cdot 1 + 1$

Raise $1$ to the power of $3$ .

$2 \cdot 1 - 5 \cdot 1^{2} + 2 \cdot 1 + 1$

Multiply $2$ by $1$ .

$2 - 5 \cdot 1^{2} + 2 \cdot 1 + 1$

Raise $1$ to the power of $2$ .

$2 - 5 \cdot 1 + 2 \cdot 1 + 1$

Multiply $- 5$ by $1$ .

$2 - 5 + 2 \cdot 1 + 1$

Subtract $5$ from $2$ .

$- 3 + 2 \cdot 1 + 1$

Multiply $2$ by $1$ .

$- 3 + 2 + 1$

Add $- 3$ and $2$ .

$- 1 + 1$

Add $- 1$ and $1$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} - 5 x^{2} + 2 x + 1}{x - 1}$

Divide $2 x^{3} - 5 x^{2} + 2 x + 1$ by $x - 1$ .

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

$x$

$1$

$2 x^{3}$

$5 x^{2}$

$2 x$

$1$

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

					$2 x^{2}$
	$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			+	$2 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$

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

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$

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

				$2 x^{2}$	-	$3 x$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$3 x$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					-	$3 x^{2}$	+	$3 x$

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

				$2 x^{2}$	-	$3 x$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					+	$3 x^{2}$	-	$3 x$

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

				$2 x^{2}$	-	$3 x$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					+	$3 x^{2}$	-	$3 x$
							-	$x$

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

				$2 x^{2}$	-	$3 x$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					+	$3 x^{2}$	-	$3 x$
							-	$x$	+	$1$

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

				$2 x^{2}$	-	$3 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					+	$3 x^{2}$	-	$3 x$
							-	$x$	+	$1$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$3 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					+	$3 x^{2}$	-	$3 x$
							-	$x$	+	$1$
							-	$x$	+	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $- x + 1$

				$2 x^{2}$	-	$3 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	-	$5 x^{2}$	+	$2 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$3 x^{2}$	+	$2 x$
					+	$3 x^{2}$	-	$3 x$
							-	$x$	+	$1$
							+	$x$	-	$1$

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

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

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

$2 x^{2} - 3 x - 1$

Write $2 x^{3} - 5 x^{2} + 2 x + 1$ as a set of factors.

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

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

$x - 1 = 0$

$2 x^{2} - 3 x - 1 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $2 x^{2} - 3 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x^{2} - 3 x - 1$ equal to $0$ .

$2 x^{2} - 3 x - 1 = 0$

Solve $2 x^{2} - 3 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 3$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 1)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 1$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 1}}{2 \cdot 2}$

Multiply $- 8$ by $- 1$ .

$x = \frac{3 \pm \sqrt{9 + 8}}{2 \cdot 2}$

Add $9$ and $8$ .

$x = \frac{3 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{17}}{4}$

The final answer is the combination of both solutions.

$x = \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4}$

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

$x = 1, \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 1, \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4}$

Decimal Form:

$x = 1, 1.78077640 \dots, - 0.28077640 \dots$

Step 4

Find the Roots (Zeros) 2x^3-5x^2+2x+1

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