Find the Roots (Zeros) 12x^3-91x^2+50x-7=0

$12 x^{3} - 91 x^{2} + 50 x - 7 = 0$

Step 1

Factor the left side of the equation.

Factor $12 x^{3} - 91 x^{2} + 50 x - 7$ 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.

$p = \pm 1, \pm 7 q = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

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

$\pm 1, \pm 0.08\overline{3}, \pm 0.5, \pm 0.1\overline{6}, \pm 0.\overline{3}, \pm 0.25, \pm 7, \pm 0.58\overline{3}, \pm 3.5, \pm 1.1\overline{6}, \pm 2.\overline{3}, \pm 1.75$

Substitute $0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $0.\overline{3}$ is a root of the polynomial.

Substitute $0.\overline{3}$ into the polynomial.

$12 \cdot {0.\overline{3}}^{3} - 91 \cdot {0.\overline{3}}^{2} + 50 \cdot 0.\overline{3} - 7$

Raise $0.\overline{3}$ to the power of $3$ .

$12 \cdot 0.\overline{037} - 91 \cdot {0.\overline{3}}^{2} + 50 \cdot 0.\overline{3} - 7$

Multiply $12$ by $0.\overline{037}$ .

$0.\overline{4} - 91 \cdot {0.\overline{3}}^{2} + 50 \cdot 0.\overline{3} - 7$

Raise $0.\overline{3}$ to the power of $2$ .

$0.\overline{4} - 91 \cdot 0.\overline{1} + 50 \cdot 0.\overline{3} - 7$

Multiply $- 91$ by $0.\overline{1}$ .

$0.\overline{4} - 10.11111111 + 50 \cdot 0.\overline{3} - 7$

Subtract $10.11111111$ from $0.\overline{4}$ .

$- 9.66666666 + 50 \cdot 0.\overline{3} - 7$

Multiply $50$ by $0.\overline{3}$ .

$- 9.66666666 + 16.\overline{6} - 7$

Add $- 9.66666666$ and $16.\overline{6}$ .

$7 - 7$

Subtract $7$ from $7$ .

$0$

Since $0.\overline{3}$ is a known root, divide the polynomial by $3 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{12 x^{3} - 91 x^{2} + 50 x - 7}{3 x - 1}$

Divide $12 x^{3} - 91 x^{2} + 50 x - 7$ by $3 x - 1$ .

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

$3 x$

$1$

$12 x^{3}$

$91 x^{2}$

$50 x$

$7$

Divide the highest order term in the dividend $12 x^{3}$ by the highest order term in divisor $3 x$ .

					$4 x^{2}$
	$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			+	$12 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{3} - 4 x^{2}$

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$

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

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$

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

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$

Divide the highest order term in the dividend $- 87 x^{2}$ by the highest order term in divisor $3 x$ .

				$4 x^{2}$	-	$29 x$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$29 x$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					-	$87 x^{2}$	+	$29 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 87 x^{2} + 29 x$

				$4 x^{2}$	-	$29 x$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$

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

				$4 x^{2}$	-	$29 x$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$
							+	$21 x$

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

				$4 x^{2}$	-	$29 x$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$
							+	$21 x$	-	$7$

Divide the highest order term in the dividend $21 x$ by the highest order term in divisor $3 x$ .

				$4 x^{2}$	-	$29 x$	+	$7$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$
							+	$21 x$	-	$7$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$29 x$	+	$7$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$
							+	$21 x$	-	$7$
							+	$21 x$	-	$7$

The expression needs to be subtracted from the dividend, so change all the signs in $21 x - 7$

				$4 x^{2}$	-	$29 x$	+	$7$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$
							+	$21 x$	-	$7$
							-	$21 x$	+	$7$

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

				$4 x^{2}$	-	$29 x$	+	$7$
$3 x$	-	$1$		$12 x^{3}$	-	$91 x^{2}$	+	$50 x$	-	$7$
			-	$12 x^{3}$	+	$4 x^{2}$
					-	$87 x^{2}$	+	$50 x$
					+	$87 x^{2}$	-	$29 x$
							+	$21 x$	-	$7$
							-	$21 x$	+	$7$
										$0$

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

$4 x^{2} - 29 x + 7$

Write $12 x^{3} - 91 x^{2} + 50 x - 7$ as a set of factors.

$(3 x - 1) (4 x^{2} - 29 x + 7) = 0$

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 $a \cdot c = 4 \cdot 7 = 28$ and whose sum is $b = - 29$ .

Factor $- 29$ out of $- 29 x$ .

$(3 x - 1) (4 x^{2} - 29 x + 7) = 0$

Rewrite $- 29$ as $- 1$ plus $- 28$

$(3 x - 1) (4 x^{2} + (- 1 - 28) x + 7) = 0$

Apply the distributive property.

$(3 x - 1) (4 x^{2} - 1 x - 28 x + 7) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x - 1) ((4 x^{2} - 1 x) - 28 x + 7) = 0$

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

$(3 x - 1) (x (4 x - 1) - 7 (4 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(3 x - 1) ((4 x - 1) (x - 7)) = 0$

Remove unnecessary parentheses.

$(3 x - 1) (4 x - 1) (x - 7) = 0$

Step 2

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

$3 x - 1 = 0$

$4 x - 1 = 0$

$x - 7 = 0$

Step 3

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 4

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 5

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 6

The final solution is all the values that make $(3 x - 1) (4 x - 1) (x - 7) = 0$ true.

$x = \frac{1}{3}, \frac{1}{4}, 7$

Step 7

Find the Roots (Zeros) 12x^3-91x^2+50x-7=0

Related Questions

Support LCMGCF.com by disabling your adblocker.