Precalculus Examples

$6 x^{3} - 43 x^{2} + 6 x + 7$

Step 1

Set $6 x^{3} - 43 x^{2} + 6 x + 7$ equal to $0$ .

$6 x^{3} - 43 x^{2} + 6 x + 7 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $6 x^{3} - 43 x^{2} + 6 x + 7$ using the rational roots test.

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Step 2.1.1.1

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.

$p = \pm 1, \pm 7$

$q = \pm 1, \pm 6, \pm 2, \pm 3$

Step 2.1.1.2

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

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

Step 2.1.1.3

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.

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Step 2.1.1.3.1

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

$6 {(- 0.\overline{3})}^{3} - 43 {(- 0.\overline{3})}^{2} + 6 \cdot - 0.\overline{3} + 7$

Step 2.1.1.3.2

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

$6 \cdot - 0.\overline{037} - 43 {(- 0.\overline{3})}^{2} + 6 \cdot - 0.\overline{3} + 7$

Step 2.1.1.3.3

Multiply $6$ by $- 0.\overline{037}$ .

$- 0.\overline{2} - 43 {(- 0.\overline{3})}^{2} + 6 \cdot - 0.\overline{3} + 7$

Step 2.1.1.3.4

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

$- 0.\overline{2} - 43 \cdot 0.\overline{1} + 6 \cdot - 0.\overline{3} + 7$

Step 2.1.1.3.5

Multiply $- 43$ by $0.\overline{1}$ .

$- 0.\overline{2} - 4.\overline{7} + 6 \cdot - 0.\overline{3} + 7$

Step 2.1.1.3.6

Subtract $4.\overline{7}$ from $- 0.\overline{2}$ .

$- 5 + 6 \cdot - 0.\overline{3} + 7$

Step 2.1.1.3.7

Multiply $6$ by $- 0.\overline{3}$ .

$- 5 - 2 + 7$

Step 2.1.1.3.8

Subtract $2$ from $- 5$ .

$- 7 + 7$

Step 2.1.1.3.9

Add $- 7$ and $7$ .

$0$

Step 2.1.1.4

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.

$\frac{6 x^{3} - 43 x^{2} + 6 x + 7}{3 x + 1}$

Step 2.1.1.5

Divide $6 x^{3} - 43 x^{2} + 6 x + 7$ by $3 x + 1$ .

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Step 2.1.1.5.1

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

$3 x$

$1$

$6 x^{3}$

$43 x^{2}$

$6 x$

$7$

Step 2.1.1.5.2

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $3 x$ .

					$2 x^{2}$
	$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$

Step 2.1.1.5.3

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			+	$6 x^{3}$	+	$2 x^{2}$

Step 2.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} + 2 x^{2}$

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$

Step 2.1.1.5.5

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$

Step 2.1.1.5.6

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$

Step 2.1.1.5.7

Divide the highest order term in the dividend $- 45 x^{2}$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$15 x$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$

Step 2.1.1.5.8

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$15 x$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					-	$45 x^{2}$	-	$15 x$

Step 2.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 45 x^{2} - 15 x$

				$2 x^{2}$	-	$15 x$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$

Step 2.1.1.5.10

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

				$2 x^{2}$	-	$15 x$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$
							+	$21 x$

Step 2.1.1.5.11

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

				$2 x^{2}$	-	$15 x$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$
							+	$21 x$	+	$7$

Step 2.1.1.5.12

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

				$2 x^{2}$	-	$15 x$	+	$7$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$
							+	$21 x$	+	$7$

Step 2.1.1.5.13

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$15 x$	+	$7$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$
							+	$21 x$	+	$7$
							+	$21 x$	+	$7$

Step 2.1.1.5.14

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

				$2 x^{2}$	-	$15 x$	+	$7$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$
							+	$21 x$	+	$7$
							-	$21 x$	-	$7$

Step 2.1.1.5.15

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

				$2 x^{2}$	-	$15 x$	+	$7$
$3 x$	+	$1$		$6 x^{3}$	-	$43 x^{2}$	+	$6 x$	+	$7$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$45 x^{2}$	+	$6 x$
					+	$45 x^{2}$	+	$15 x$
							+	$21 x$	+	$7$
							-	$21 x$	-	$7$
										$0$

Step 2.1.1.5.16

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

$2 x^{2} - 15 x + 7$

Step 2.1.1.6

Write $6 x^{3} - 43 x^{2} + 6 x + 7$ as a set of factors.

$(3 x + 1) (2 x^{2} - 15 x + 7) = 0$

Step 2.1.2

Factor by grouping.

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Step 2.1.2.1

Factor by grouping.

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Step 2.1.2.1.1

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 $a \cdot c = 2 \cdot 7 = 14$ and whose sum is $b = - 15$ .

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Step 2.1.2.1.1.1

Factor $- 15$ out of $- 15 x$ .

$(3 x + 1) (2 x^{2} - 15 x + 7) = 0$

Step 2.1.2.1.1.2

Rewrite $- 15$ as $- 1$ plus $- 14$

$(3 x + 1) (2 x^{2} + (- 1 - 14) x + 7) = 0$

Step 2.1.2.1.1.3

Apply the distributive property.

$(3 x + 1) (2 x^{2} - 1 x - 14 x + 7) = 0$

Step 2.1.2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1.2.1

Group the first two terms and the last two terms.

$(3 x + 1) ((2 x^{2} - 1 x) - 14 x + 7) = 0$

Step 2.1.2.1.2.2

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

$(3 x + 1) (x (2 x - 1) - 7 (2 x - 1)) = 0$

Step 2.1.2.1.3

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

$(3 x + 1) ((2 x - 1) (x - 7)) = 0$

Step 2.1.2.2

Remove unnecessary parentheses.

$(3 x + 1) (2 x - 1) (x - 7) = 0$

Step 2.2

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

$3 x + 1 = 0$

$2 x - 1 = 0$

$x - 7 = 0$

Step 2.3

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

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Step 2.3.1

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

$3 x + 1 = 0$

Step 2.3.2

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

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Step 2.3.2.2

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

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

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

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.4

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

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Step 2.4.1

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

$2 x - 1 = 0$

Step 2.4.2

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

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5

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

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Step 2.5.1

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

$x - 7 = 0$

Step 2.5.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.6

The final solution is all the values that make $(3 x + 1) (2 x - 1) (x - 7) = 0$ true.

$x = - \frac{1}{3}, \frac{1}{2}, 7$

Step 3