Precalculus Examples

$3 x^{4} - 10 x^{3} - 9 x^{2} + 40 x - 12$

Step 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.

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

$q = \pm 1, \pm 3$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 4, \pm \frac{4}{3}, \pm 6, \pm 12$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$3 {(2)}^{4} - 10 {(2)}^{3} - 9 {(2)}^{2} + 40 (2) - 12$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Tap for more steps...

Step 4.1

Simplify each term.

Tap for more steps...

Step 4.1.1

Raise $2$ to the power of $4$ .

$3 \cdot 16 - 10 {(2)}^{3} - 9 {(2)}^{2} + 40 (2) - 12$

Step 4.1.2

Multiply $3$ by $16$ .

$48 - 10 {(2)}^{3} - 9 {(2)}^{2} + 40 (2) - 12$

Step 4.1.3

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

$48 - 10 \cdot 8 - 9 {(2)}^{2} + 40 (2) - 12$

Step 4.1.4

Multiply $- 10$ by $8$ .

$48 - 80 - 9 {(2)}^{2} + 40 (2) - 12$

Step 4.1.5

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

$48 - 80 - 9 \cdot 4 + 40 (2) - 12$

Step 4.1.6

Multiply $- 9$ by $4$ .

$48 - 80 - 36 + 40 (2) - 12$

Step 4.1.7

Multiply $40$ by $2$ .

$48 - 80 - 36 + 80 - 12$

Step 4.2

Simplify by adding and subtracting.

Tap for more steps...

Step 4.2.1

Subtract $80$ from $48$ .

$- 32 - 36 + 80 - 12$

Step 4.2.2

Subtract $36$ from $- 32$ .

$- 68 + 80 - 12$

Step 4.2.3

Add $- 68$ and $80$ .

$12 - 12$

Step 4.2.4

Subtract $12$ from $12$ .

$0$

Step 5

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{3 x^{4} - 10 x^{3} - 9 x^{2} + 40 x - 12}{x - 2}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Tap for more steps...

Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$

Step 6.2

The first number in the dividend $(3)$ is put into the first position of the result area (below the horizontal line).

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$

	$3$

Step 6.3

Multiply the newest entry in the result $(3)$ by the divisor $(2)$ and place the result of $(6)$ under the next term in the dividend $(- 10)$ .

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$
	$3$

Step 6.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$
	$3$	$- 4$

Step 6.5

Multiply the newest entry in the result $(- 4)$ by the divisor $(2)$ and place the result of $(- 8)$ under the next term in the dividend $(- 9)$ .

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$	$- 8$
	$3$	$- 4$

Step 6.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$	$- 8$
	$3$	$- 4$	$- 17$

Step 6.7

Multiply the newest entry in the result $(- 17)$ by the divisor $(2)$ and place the result of $(- 34)$ under the next term in the dividend $(40)$ .

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$	$- 8$	$- 34$
	$3$	$- 4$	$- 17$

Step 6.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$	$- 8$	$- 34$
	$3$	$- 4$	$- 17$	$6$

Step 6.9

Multiply the newest entry in the result $(6)$ by the divisor $(2)$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$	$- 8$	$- 34$	$12$
	$3$	$- 4$	$- 17$	$6$

Step 6.10

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$- 10$	$- 9$	$40$	$- 12$
		$6$	$- 8$	$- 34$	$12$
	$3$	$- 4$	$- 17$	$6$	$0$

Step 6.11

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$3 x^{3} + - 4 x^{2} + (- 17) x + 6$

Step 6.12

Simplify the quotient polynomial.

$3 x^{3} - 4 x^{2} - 17 x + 6$

Step 7

Solve the equation to find any remaining roots.

Tap for more steps...

Step 7.1

Factor the left side of the equation.

Tap for more steps...

Step 7.1.1

Factor $3 x^{3} - 4 x^{2} - 17 x + 6$ using the rational roots test.

Tap for more steps...

Step 7.1.1.1

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.

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

$q = \pm 1, \pm 3$

Step 7.1.1.2

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

$\pm 1, \pm 0.\overline{3}, \pm 6, \pm 2, \pm 0.\overline{6}, \pm 3$

Step 7.1.1.3

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.

Tap for more steps...

Step 7.1.1.3.1

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

$3 \cdot {0.\overline{3}}^{3} - 4 \cdot {0.\overline{3}}^{2} - 17 \cdot 0.\overline{3} + 6$

Step 7.1.1.3.2

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

$3 \cdot 0.\overline{037} - 4 \cdot {0.\overline{3}}^{2} - 17 \cdot 0.\overline{3} + 6$

Step 7.1.1.3.3

Multiply $3$ by $0.\overline{037}$ .

$0.\overline{1} - 4 \cdot {0.\overline{3}}^{2} - 17 \cdot 0.\overline{3} + 6$

Step 7.1.1.3.4

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

$0.\overline{1} - 4 \cdot 0.\overline{1} - 17 \cdot 0.\overline{3} + 6$

Step 7.1.1.3.5

Multiply $- 4$ by $0.\overline{1}$ .

$0.\overline{1} - 0.\overline{4} - 17 \cdot 0.\overline{3} + 6$

Step 7.1.1.3.6

Subtract $0.\overline{4}$ from $0.\overline{1}$ .

$- 0.\overline{3} - 17 \cdot 0.\overline{3} + 6$

Step 7.1.1.3.7

Multiply $- 17$ by $0.\overline{3}$ .

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

Step 7.1.1.3.8

Subtract $5.\overline{6}$ from $- 0.\overline{3}$ .

$- 6 + 6$

Step 7.1.1.3.9

Add $- 6$ and $6$ .

$0$

Step 7.1.1.4

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.

$\frac{3 x^{3} - 4 x^{2} - 17 x + 6}{3 x - 1}$

Step 7.1.1.5

Divide $3 x^{3} - 4 x^{2} - 17 x + 6$ by $3 x - 1$ .

Tap for more steps...

Step 7.1.1.5.1

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

$3 x$

$1$

$3 x^{3}$

$4 x^{2}$

$17 x$

$6$

Step 7.1.1.5.2

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

					$x^{2}$
	$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$

Step 7.1.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			+	$3 x^{3}$	-	$x^{2}$

Step 7.1.1.5.4

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$

Step 7.1.1.5.5

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$

Step 7.1.1.5.6

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$

Step 7.1.1.5.7

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$

Step 7.1.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					-	$3 x^{2}$	+	$x$

Step 7.1.1.5.9

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$

Step 7.1.1.5.10

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$

Step 7.1.1.5.11

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

				$x^{2}$	-	$x$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$

Step 7.1.1.5.12

Divide the highest order term in the dividend $- 18 x$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$

Step 7.1.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$
							-	$18 x$	+	$6$

Step 7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 18 x + 6$

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$
							+	$18 x$	-	$6$

Step 7.1.1.5.15

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

				$x^{2}$	-	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	-	$4 x^{2}$	-	$17 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					-	$3 x^{2}$	-	$17 x$
					+	$3 x^{2}$	-	$x$
							-	$18 x$	+	$6$
							+	$18 x$	-	$6$
										$0$

Step 7.1.1.5.16

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

$x^{2} - x - 6$

Step 7.1.1.6

Write $3 x^{3} - 4 x^{2} - 17 x + 6$ as a set of factors.

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

Step 7.1.2

Factor $x^{2} - x - 6$ using the AC method.

Tap for more steps...

Step 7.1.2.1

Factor $x^{2} - x - 6$ using the AC method.

Tap for more steps...

Step 7.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 7.1.2.1.2

Write the factored form using these integers.

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

Step 7.1.2.2

Remove unnecessary parentheses.

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

Step 7.2

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

$3 x - 1 = 0$

$x - 3 = 0$

$x + 2 = 0$

Step 7.3

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

Tap for more steps...

Step 7.3.1

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

$3 x - 1 = 0$

Step 7.3.2

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

Tap for more steps...

Step 7.3.2.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 7.3.2.2

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

Tap for more steps...

Step 7.3.2.2.1

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

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

Step 7.3.2.2.2

Simplify the left side.

Tap for more steps...

Step 7.3.2.2.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 7.3.2.2.2.1.1

Cancel the common factor.

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

Step 7.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 7.4

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

Tap for more steps...

Step 7.4.1

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

$x - 3 = 0$

Step 7.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7.5

Set $x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 7.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 7.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7.6

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

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

Step 8

The polynomial can be written as a set of linear factors.

$(x - 2) (x - \frac{1}{3}) (x - 3) (x + 2)$

Step 9

These are the roots (zeros) of the polynomial $3 x^{4} - 10 x^{3} - 9 x^{2} + 40 x - 12$ .

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

Step 10