Finite Math Examples

$4 r^{2} + 20 r + 25$

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.

$p = \pm 1, \pm 5, \pm 25$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 25, \pm \frac{25}{2}, \pm \frac{25}{4}$

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.

$4 {(- \frac{5}{2})}^{2} + 20 (- \frac{5}{2}) + 25$

Step 4

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

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Step 4.1

Simplify each term.

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Step 4.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.1.1.1

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

$4 ({(- 1)}^{2} {(\frac{5}{2})}^{2}) + 20 (- \frac{5}{2}) + 25$

Step 4.1.1.2

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

$4 ({(- 1)}^{2} \frac{5^{2}}{2^{2}}) + 20 (- \frac{5}{2}) + 25$

Step 4.1.2

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

$4 (1 \frac{5^{2}}{2^{2}}) + 20 (- \frac{5}{2}) + 25$

Step 4.1.3

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$4 \frac{5^{2}}{2^{2}} + 20 (- \frac{5}{2}) + 25$

Step 4.1.4

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

$4 \frac{25}{2^{2}} + 20 (- \frac{5}{2}) + 25$

Step 4.1.5

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

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

Step 4.1.6

Cancel the common factor of $4$ .

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Step 4.1.6.1

Cancel the common factor.

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

Step 4.1.6.2

Rewrite the expression.

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

Step 4.1.7

Cancel the common factor of $2$ .

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Step 4.1.7.1

Move the leading negative in $- \frac{5}{2}$ into the numerator.

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

Step 4.1.7.2

Factor $2$ out of $20$ .

$25 + 2 (10) \frac{- 5}{2} + 25$

Step 4.1.7.3

Cancel the common factor.

$25 + 2 \cdot 10 \frac{- 5}{2} + 25$

Step 4.1.7.4

Rewrite the expression.

$25 + 10 \cdot - 5 + 25$

Step 4.1.8

Multiply $10$ by $- 5$ .

$25 - 50 + 25$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $50$ from $25$ .

$- 25 + 25$

Step 4.2.2

Add $- 25$ and $25$ .

$0$

Step 5

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

$\frac{4 r^{2} + 20 r + 25}{r + \frac{5}{2}}$

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.

$- \frac{5}{2}$	$4$	$20$	$25$

Step 6.2

The first number in the dividend $(4)$ is put into the first position of the result area (below the horizontal line).

$- \frac{5}{2}$	$4$	$20$	$25$

	$4$

Step 6.3

Multiply the newest entry in the result $(4)$ by the divisor $(- \frac{5}{2})$ and place the result of $(- 10)$ under the next term in the dividend $(20)$ .

$- \frac{5}{2}$	$4$	$20$	$25$
		$- 10$
	$4$

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.

$- \frac{5}{2}$	$4$	$20$	$25$
		$- 10$
	$4$	$10$

Step 6.5

Multiply the newest entry in the result $(10)$ by the divisor $(- \frac{5}{2})$ and place the result of $(- 25)$ under the next term in the dividend $(25)$ .

$- \frac{5}{2}$	$4$	$20$	$25$
		$- 10$	$- 25$
	$4$	$10$

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.

$- \frac{5}{2}$	$4$	$20$	$25$
		$- 10$	$- 25$
	$4$	$10$	$0$

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.

$(4) r + 10$

Step 6.8

Simplify the quotient polynomial.

$4 r + 10$

Step 7

Factor $2$ out of $4 r + 10$ .

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Step 7.1

Factor $2$ out of $4 r$ .

$(r + \frac{5}{2}) (2 (2 r) + 10)$

Step 7.2

Factor $2$ out of $10$ .

$(r + \frac{5}{2}) (2 (2 r) + 2 (5))$

Step 7.3

Factor $2$ out of $2 (2 r) + 2 (5)$ .

$(r + \frac{5}{2}) (2 (2 r + 5))$

Step 8

Factor using the perfect square rule.

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Step 8.1

Rewrite $4 r^{2}$ as ${(2 r)}^{2}$ .

${(2 r)}^{2} + 20 r + 25 = 0$

Step 8.2

Rewrite $25$ as $5^{2}$ .

${(2 r)}^{2} + 20 r + 5^{2} = 0$

Step 8.3

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

$20 r = 2 \cdot (2 r) \cdot 5$

Step 8.4

Rewrite the polynomial.

${(2 r)}^{2} + 2 \cdot (2 r) \cdot 5 + 5^{2} = 0$

Step 8.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 r$ and $b = 5$ .

${(2 r + 5)}^{2} = 0$

Step 9

Set the $2 r + 5$ equal to $0$ .

$2 r + 5 = 0$

Step 10

Solve for $r$ .

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Step 10.1

Subtract $5$ from both sides of the equation.

$2 r = - 5$

Step 10.2

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

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Step 10.2.1

Divide each term in $2 r = - 5$ by $2$ .

$\frac{2 r}{2} = \frac{- 5}{2}$

Step 10.2.2

Simplify the left side.

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Step 10.2.2.1

Cancel the common factor of $2$ .

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Step 10.2.2.1.1

Cancel the common factor.

$\frac{2 r}{2} = \frac{- 5}{2}$

Step 10.2.2.1.2

Divide $r$ by $1$ .

$r = \frac{- 5}{2}$

Step 10.2.3

Simplify the right side.

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Step 10.2.3.1

Move the negative in front of the fraction.

$r = - \frac{5}{2}$

Step 11