Find the Roots (Zeros) 2x^4-7x^3+3x^2+8x-4

$2 x^{4} - 7 x^{3} + 3 x^{2} + 8 x - 4$

Step 1

Set $2 x^{4} - 7 x^{3} + 3 x^{2} + 8 x - 4$ equal to $0$ .

$2 x^{4} - 7 x^{3} + 3 x^{2} + 8 x - 4 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$8 x - 4 + 2 x^{4} - 7 x^{3} + 3 x^{2} = 0$

Factor $4$ out of $8 x - 4$ .

Factor $4$ out of $8 x$ .

$4 (2 x) - 4 + 2 x^{4} - 7 x^{3} + 3 x^{2} = 0$

Factor $4$ out of $- 4$ .

$4 (2 x) + 4 (- 1) + 2 x^{4} - 7 x^{3} + 3 x^{2} = 0$

Factor $4$ out of $4 (2 x) + 4 (- 1)$ .

$4 (2 x - 1) + 2 x^{4} - 7 x^{3} + 3 x^{2} = 0$

Factor $x^{2}$ out of $2 x^{4} - 7 x^{3} + 3 x^{2}$ .

Factor $x^{2}$ out of $2 x^{4}$ .

$4 (2 x - 1) + x^{2} (2 x^{2}) - 7 x^{3} + 3 x^{2} = 0$

Factor $x^{2}$ out of $- 7 x^{3}$ .

$4 (2 x - 1) + x^{2} (2 x^{2}) + x^{2} (- 7 x) + 3 x^{2} = 0$

Factor $x^{2}$ out of $3 x^{2}$ .

$4 (2 x - 1) + x^{2} (2 x^{2}) + x^{2} (- 7 x) + x^{2} \cdot 3 = 0$

Factor $x^{2}$ out of $x^{2} (2 x^{2}) + x^{2} (- 7 x)$ .

$4 (2 x - 1) + x^{2} (2 x^{2} - 7 x) + x^{2} \cdot 3 = 0$

Factor $x^{2}$ out of $x^{2} (2 x^{2} - 7 x) + x^{2} \cdot 3$ .

$4 (2 x - 1) + x^{2} (2 x^{2} - 7 x + 3) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$4 (2 x - 1) + x^{2} (2 x^{2} - 7 x + 3) = 0$

Rewrite $- 7$ as $- 1$ plus $- 6$

$4 (2 x - 1) + x^{2} (2 x^{2} + (- 1 - 6) x + 3) = 0$

Apply the distributive property.

$4 (2 x - 1) + x^{2} (2 x^{2} - 1 x - 6 x + 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$4 (2 x - 1) + x^{2} ((2 x^{2} - 1 x) - 6 x + 3) = 0$

Factor out the greatest common factor (GCF) from each group.

$4 (2 x - 1) + x^{2} (x (2 x - 1) - 3 (2 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$4 (2 x - 1) + x^{2} ((2 x - 1) (x - 3)) = 0$

Remove unnecessary parentheses.

$4 (2 x - 1) + x^{2} (2 x - 1) (x - 3) = 0$

Factor $2 x - 1$ out of $4 (2 x - 1) + x^{2} (2 x - 1) (x - 3)$ .

Factor $2 x - 1$ out of $4 (2 x - 1)$ .

$(2 x - 1) \cdot 4 + x^{2} (2 x - 1) (x - 3) = 0$

Factor $2 x - 1$ out of $x^{2} (2 x - 1) (x - 3)$ .

$(2 x - 1) \cdot 4 + (2 x - 1) (x^{2} (x - 3)) = 0$

Factor $2 x - 1$ out of $(2 x - 1) \cdot 4 + (2 x - 1) (x^{2} (x - 3))$ .

$(2 x - 1) (4 + x^{2} (x - 3)) = 0$

Apply the distributive property.

$(2 x - 1) (4 + x^{2} x + x^{2} \cdot - 3) = 0$

Multiply $x^{2}$ by $x$ by adding the exponents.

Multiply $x^{2}$ by $x$ .

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

$(2 x - 1) (4 + x^{2} x + x^{2} \cdot - 3) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(2 x - 1) (4 + x^{2 + 1} + x^{2} \cdot - 3) = 0$

Add $2$ and $1$ .

$(2 x - 1) (4 + x^{3} + x^{2} \cdot - 3) = 0$

Move $- 3$ to the left of $x^{2}$ .

$(2 x - 1) (4 + x^{3} - 3 x^{2}) = 0$

Reorder terms.

$(2 x - 1) (x^{3} - 3 x^{2} + 4) = 0$

Factor.

Rewrite $x^{3} - 3 x^{2} + 4$ in a factored form.

Factor $x^{3} - 3 x^{2} + 4$ 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.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1$

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

$\pm 1, \pm 4, \pm 2$

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.

${(- 1)}^{3} - 3 {(- 1)}^{2} + 4$

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

$- 1 - 3 {(- 1)}^{2} + 4$

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

$- 1 - 3 \cdot 1 + 4$

Multiply $- 3$ by $1$ .

$- 1 - 3 + 4$

Subtract $3$ from $- 1$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

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.

$\frac{x^{3} - 3 x^{2} + 4}{x + 1}$

Divide $x^{3} - 3 x^{2} + 4$ 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 $0$ .

$x$

$1$

$x^{3}$

$3 x^{2}$

$0 x$

$4$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					-	$4 x^{2}$	-	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} - 4 x$

				$x^{2}$	-	$4 x$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$

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

				$x^{2}$	-	$4 x$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$
							+	$4 x$

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

				$x^{2}$	-	$4 x$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$
							+	$4 x$	+	$4$

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$	+	$4$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$
							+	$4 x$	+	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	+	$4$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$
							+	$4 x$	+	$4$
							+	$4 x$	+	$4$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x + 4$

				$x^{2}$	-	$4 x$	+	$4$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$
							+	$4 x$	+	$4$
							-	$4 x$	-	$4$

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

				$x^{2}$	-	$4 x$	+	$4$
$x$	+	$1$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					-	$4 x^{2}$	+	$0 x$
					+	$4 x^{2}$	+	$4 x$
							+	$4 x$	+	$4$
							-	$4 x$	-	$4$
										$0$

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

$x^{2} - 4 x + 4$

Write $x^{3} - 3 x^{2} + 4$ as a set of factors.

$(2 x - 1) ((x + 1) (x^{2} - 4 x + 4)) = 0$

Factor using the perfect square rule.

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

$(2 x - 1) ((x + 1) (x^{2} - 4 x + 2^{2})) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$(2 x - 1) ((x + 1) (x^{2} - 2 \cdot x \cdot 2 + 2^{2})) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

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

$2 x - 1 = 0$

$x + 1 = 0$

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

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 2)}^{2}$ equal to $0$ .

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

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = \frac{1}{2}, - 1, 2$

Step 3

Find the Roots (Zeros) 2x^4-7x^3+3x^2+8x-4

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