Finite Math Examples

$2 x^{2} + 20 x + 48$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48$

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.

$2 {(- 4)}^{2} + 20 (- 4) + 48$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 4$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

$2 \cdot 16 + 20 (- 4) + 48$

Step 4.1.2

Multiply $2$ by $16$ .

$32 + 20 (- 4) + 48$

Step 4.1.3

Multiply $20$ by $- 4$ .

$32 - 80 + 48$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $80$ from $32$ .

$- 48 + 48$

Step 4.2.2

Add $- 48$ and $48$ .

$0$

Step 5

Since $- 4$ is a known root, divide the polynomial by $x + 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{2} + 20 x + 48}{x + 4}$

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.

$- 4$	$2$	$20$	$48$

Step 6.2

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$- 4$	$2$	$20$	$48$

	$2$

Step 6.3

Multiply the newest entry in the result $(2)$ by the divisor $(- 4)$ and place the result of $(- 8)$ under the next term in the dividend $(20)$ .

$- 4$	$2$	$20$	$48$
		$- 8$
	$2$

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.

$- 4$	$2$	$20$	$48$
		$- 8$
	$2$	$12$

Step 6.5

Multiply the newest entry in the result $(12)$ by the divisor $(- 4)$ and place the result of $(- 48)$ under the next term in the dividend $(48)$ .

$- 4$	$2$	$20$	$48$
		$- 8$	$- 48$
	$2$	$12$

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.

$- 4$	$2$	$20$	$48$
		$- 8$	$- 48$
	$2$	$12$	$0$

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.

$(2) x + 12$

Step 6.8

Simplify the quotient polynomial.

$2 x + 12$

Step 7

Factor $2$ out of $2 x + 12$ .

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Step 7.1

Factor $2$ out of $2 x$ .

$(x + 4) (2 (x) + 12)$

Step 7.2

Factor $2$ out of $12$ .

$(x + 4) (2 x + 2 \cdot 6)$

Step 7.3

Factor $2$ out of $2 x + 2 \cdot 6$ .

$(x + 4) (2 (x + 6))$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $2$ out of $2 x^{2} + 20 x + 48$ .

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Step 8.1.1

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 20 x + 48 = 0$

Step 8.1.2

Factor $2$ out of $20 x$ .

$2 (x^{2}) + 2 (10 x) + 48 = 0$

Step 8.1.3

Factor $2$ out of $48$ .

$2 x^{2} + 2 (10 x) + 2 \cdot 24 = 0$

Step 8.1.4

Factor $2$ out of $2 x^{2} + 2 (10 x)$ .

$2 (x^{2} + 10 x) + 2 \cdot 24 = 0$

Step 8.1.5

Factor $2$ out of $2 (x^{2} + 10 x) + 2 \cdot 24$ .

$2 (x^{2} + 10 x + 24) = 0$

Step 8.2

Factor.

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Step 8.2.1

Factor $x^{2} + 10 x + 24$ using the AC method.

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Step 8.2.1.1

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 $24$ and whose sum is $10$ .

$4, 6$

Step 8.2.1.2

Write the factored form using these integers.

$2 ((x + 4) (x + 6)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$2 (x + 4) (x + 6) = 0$

Step 9

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

$x + 4 = 0$

$x + 6 = 0$

Step 10

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

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Step 10.1

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

$x + 4 = 0$

Step 10.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 11

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 11.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 12

The final solution is all the values that make $2 (x + 4) (x + 6) = 0$ true.

$x = - 4, - 6$

Step 13