Precalculus Examples

$6 x^{2} - 7 x - 20$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}, \pm 10, \pm \frac{10}{3}, \pm 20, \pm \frac{20}{3}$

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.

$6 {(\frac{5}{2})}^{2} - 7 (\frac{5}{2}) - 20$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{5}{2}$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

Apply the product rule to $\frac{5}{2}$ .

$6 \frac{5^{2}}{2^{2}} - 7 (\frac{5}{2}) - 20$

Step 4.1.2

Raise $5$ to the power of $2$ .

$6 \frac{25}{2^{2}} - 7 (\frac{5}{2}) - 20$

Step 4.1.3

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

$6 (\frac{25}{4}) - 7 (\frac{5}{2}) - 20$

Step 4.1.4

Cancel the common factor of $2$ .

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Step 4.1.4.1

Factor $2$ out of $6$ .

$2 (3) \frac{25}{4} - 7 (\frac{5}{2}) - 20$

Step 4.1.4.2

Factor $2$ out of $4$ .

$2 \cdot 3 \frac{25}{2 \cdot 2} - 7 (\frac{5}{2}) - 20$

Step 4.1.4.3

Cancel the common factor.

$2 \cdot 3 \frac{25}{2 \cdot 2} - 7 (\frac{5}{2}) - 20$

Step 4.1.4.4

Rewrite the expression.

$3 (\frac{25}{2}) - 7 (\frac{5}{2}) - 20$

Step 4.1.5

Combine $3$ and $\frac{25}{2}$ .

$\frac{3 \cdot 25}{2} - 7 (\frac{5}{2}) - 20$

Step 4.1.6

Multiply $3$ by $25$ .

$\frac{75}{2} - 7 (\frac{5}{2}) - 20$

Step 4.1.7

Multiply $- 7 (\frac{5}{2})$ .

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Step 4.1.7.1

Combine $- 7$ and $\frac{5}{2}$ .

$\frac{75}{2} + \frac{- 7 \cdot 5}{2} - 20$

Step 4.1.7.2

Multiply $- 7$ by $5$ .

$\frac{75}{2} + \frac{- 35}{2} - 20$

Step 4.1.8

Move the negative in front of the fraction.

$\frac{75}{2} - \frac{35}{2} - 20$

Step 4.2

Combine fractions.

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Step 4.2.1

Combine the numerators over the common denominator.

$- 20 + \frac{75 - 35}{2}$

Step 4.2.2

Simplify the expression.

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Step 4.2.2.1

Subtract $35$ from $75$ .

$- 20 + \frac{40}{2}$

Step 4.2.2.2

Divide $40$ by $2$ .

$- 20 + 20$

Step 4.2.2.3

Add $- 20$ and $20$ .

$0$

Step 5

Since $\frac{5}{2}$ is a known root, divide the polynomial by $x - \frac{5}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{2} - 7 x - 20}{x - \frac{5}{2}}$

Step 6

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

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Step 6.1

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

$\frac{5}{2}$	$6$	$- 7$	$- 20$

Step 6.2

The first number in the dividend $(6)$ is put into the first position of the result area (below the horizontal line).

$\frac{5}{2}$	$6$	$- 7$	$- 20$

	$6$

Step 6.3

Multiply the newest entry in the result $(6)$ by the divisor $(\frac{5}{2})$ and place the result of $(15)$ under the next term in the dividend $(- 7)$ .

$\frac{5}{2}$	$6$	$- 7$	$- 20$
		$15$
	$6$

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.

$\frac{5}{2}$	$6$	$- 7$	$- 20$
		$15$
	$6$	$8$

Step 6.5

Multiply the newest entry in the result $(8)$ by the divisor $(\frac{5}{2})$ and place the result of $(20)$ under the next term in the dividend $(- 20)$ .

$\frac{5}{2}$	$6$	$- 7$	$- 20$
		$15$	$20$
	$6$	$8$

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.

$\frac{5}{2}$	$6$	$- 7$	$- 20$
		$15$	$20$
	$6$	$8$	$0$

Step 6.7

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

$(6) x + 8$

Step 6.8

Simplify the quotient polynomial.

$6 x + 8$

Step 7

Factor $2$ out of $6 x + 8$ .

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Step 7.1

Factor $2$ out of $6 x$ .

$(x - \frac{5}{2}) (2 (3 x) + 8)$

Step 7.2

Factor $2$ out of $8$ .

$(x - \frac{5}{2}) (2 (3 x) + 2 (4))$

Step 7.3

Factor $2$ out of $2 (3 x) + 2 (4)$ .

$(x - \frac{5}{2}) (2 (3 x + 4))$

Step 8

Factor by grouping.

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Step 8.1

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 $a \cdot c = 6 \cdot - 20 = - 120$ and whose sum is $b = - 7$ .

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Step 8.1.1

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

$6 x^{2} - 7 x - 20 = 0$

Step 8.1.2

Rewrite $- 7$ as $8$ plus $- 15$

$6 x^{2} + (8 - 15) x - 20 = 0$

Step 8.1.3

Apply the distributive property.

$6 x^{2} + 8 x - 15 x - 20 = 0$

Step 8.2

Factor out the greatest common factor from each group.

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Step 8.2.1

Group the first two terms and the last two terms.

$(6 x^{2} + 8 x) - 15 x - 20 = 0$

Step 8.2.2

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

$2 x (3 x + 4) - 5 (3 x + 4) = 0$

Step 8.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 4$ .

$(3 x + 4) (2 x - 5) = 0$

Step 9

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

$3 x + 4 = 0$

$2 x - 5 = 0$

Step 10

Set $3 x + 4$ equal to $0$ and solve for $x$ .

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Step 10.1

Set $3 x + 4$ equal to $0$ .

$3 x + 4 = 0$

Step 10.2

Solve $3 x + 4 = 0$ for $x$ .

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Step 10.2.1

Subtract $4$ from both sides of the equation.

$3 x = - 4$

Step 10.2.2

Divide each term in $3 x = - 4$ by $3$ and simplify.

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Step 10.2.2.1

Divide each term in $3 x = - 4$ by $3$ .

$\frac{3 x}{3} = \frac{- 4}{3}$

Step 10.2.2.2

Simplify the left side.

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Step 10.2.2.2.1

Cancel the common factor of $3$ .

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Step 10.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 4}{3}$

Step 10.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 4}{3}$

Step 10.2.2.3

Simplify the right side.

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Step 10.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{4}{3}$

Step 11

Set $2 x - 5$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 11.2

Solve $2 x - 5 = 0$ for $x$ .

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Step 11.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 11.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 11.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 11.2.2.2

Simplify the left side.

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Step 11.2.2.2.1

Cancel the common factor of $2$ .

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Step 11.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 11.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 12

The final solution is all the values that make $(3 x + 4) (2 x - 5) = 0$ true.

$x = - \frac{4}{3}, \frac{5}{2}$

Step 13