Algebra Examples

$f (x) = 5 x^{4} - 4 x^{3} + 44 x^{2} - 36 x - 9$

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.

$p = \pm 1, \pm 3, \pm 9$

$q = \pm 1, \pm 5$

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

$\pm 1, \pm \frac{1}{5}, \pm 3, \pm \frac{3}{5}, \pm 9, \pm \frac{9}{5}$

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.

$5 {(1)}^{4} - 4 {(1)}^{3} + 44 {(1)}^{2} - 36 \cdot 1 - 9$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Simplify each term.

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One to any power is one.

$5 \cdot 1 - 4 {(1)}^{3} + 44 {(1)}^{2} - 36 \cdot 1 - 9$

Multiply $5$ by $1$ .

$5 - 4 {(1)}^{3} + 44 {(1)}^{2} - 36 \cdot 1 - 9$

One to any power is one.

$5 - 4 \cdot 1 + 44 {(1)}^{2} - 36 \cdot 1 - 9$

Multiply $- 4$ by $1$ .

$5 - 4 + 44 {(1)}^{2} - 36 \cdot 1 - 9$

One to any power is one.

$5 - 4 + 44 \cdot 1 - 36 \cdot 1 - 9$

Multiply $44$ by $1$ .

$5 - 4 + 44 - 36 \cdot 1 - 9$

Multiply $- 36$ by $1$ .

$5 - 4 + 44 - 36 - 9$

Simplify by adding and subtracting.

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Subtract $4$ from $5$ .

$1 + 44 - 36 - 9$

Add $1$ and $44$ .

$45 - 36 - 9$

Subtract $36$ from $45$ .

$9 - 9$

Subtract $9$ from $9$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{5 x^{4} - 4 x^{3} + 44 x^{2} - 36 x - 9}{x - 1}$

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

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

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$

The first number in the dividend $(5)$ is put into the first position of the result area (below the horizontal line).

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$

	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(1)$ and place the result of $(5)$ under the next term in the dividend $(- 4)$ .

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$
	$5$

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

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$
	$5$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(44)$ .

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$	$1$
	$5$	$1$

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

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$	$1$
	$5$	$1$	$45$

Multiply the newest entry in the result $(45)$ by the divisor $(1)$ and place the result of $(45)$ under the next term in the dividend $(- 36)$ .

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$	$1$	$45$
	$5$	$1$	$45$

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

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$	$1$	$45$
	$5$	$1$	$45$	$9$

Multiply the newest entry in the result $(9)$ by the divisor $(1)$ and place the result of $(9)$ under the next term in the dividend $(- 9)$ .

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$	$1$	$45$	$9$
	$5$	$1$	$45$	$9$

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

$1$	$5$	$- 4$	$44$	$- 36$	$- 9$
		$5$	$1$	$45$	$9$
	$5$	$1$	$45$	$9$	$0$

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

$5 x^{3} + 1 x^{2} + (45) x + 9$

Simplify the quotient polynomial.

$5 x^{3} + x^{2} + 45 x + 9$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(x - 1) (5 x^{3} + x^{2}) + 45 x + 9$

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

$(x - 1) + x^{2} (5 x + 1) + 9 (5 x + 1)$

$(x - 1) x^{2} (5 x + 1) + 9 (5 x + 1)$

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

$(x - 1) (5 x + 1) (x^{2} + 9)$

Factor the left side of the equation.

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Regroup terms.

$- 4 x^{3} - 36 x + 5 x^{4} + 44 x^{2} - 9 = 0$

Factor $- 4 x$ out of $- 4 x^{3} - 36 x$ .

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Factor $- 4 x$ out of $- 4 x^{3}$ .

$- 4 x \cdot x^{2} - 36 x + 5 x^{4} + 44 x^{2} - 9 = 0$

Factor $- 4 x$ out of $- 36 x$ .

$- 4 x \cdot x^{2} - 4 x \cdot 9 + 5 x^{4} + 44 x^{2} - 9 = 0$

Factor $- 4 x$ out of $- 4 x (x^{2}) - 4 x (9)$ .

$- 4 x (x^{2} + 9) + 5 x^{4} + 44 x^{2} - 9 = 0$

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

$- 4 x (x^{2} + 9) + 5 {(x^{2})}^{2} + 44 x^{2} - 9 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 4 x (x^{2} + 9) + 5 u^{2} + 44 u - 9 = 0$

Factor by grouping.

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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 $a \cdot c = 5 \cdot - 9 = - 45$ and whose sum is $b = 44$ .

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Factor $44$ out of $44 u$ .

$- 4 x (x^{2} + 9) + 5 u^{2} + 44 (u) - 9 = 0$

Rewrite $44$ as $- 1$ plus $45$

$- 4 x (x^{2} + 9) + 5 u^{2} + (- 1 + 45) u - 9 = 0$

Apply the distributive property.

$- 4 x (x^{2} + 9) + 5 u^{2} - 1 u + 45 u - 9 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$- 4 x (x^{2} + 9) + 5 u^{2} - 1 u + 45 u - 9 = 0$

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

$- 4 x (x^{2} + 9) + u (5 u - 1) + 9 (5 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 u - 1$ .

$- 4 x (x^{2} + 9) + (5 u - 1) (u + 9) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$- 4 x (x^{2} + 9) + (5 x^{2} - 1) (x^{2} + 9) = 0$

Factor $x^{2} + 9$ out of $- 4 x (x^{2} + 9) + (5 x^{2} - 1) (x^{2} + 9)$ .

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Factor $x^{2} + 9$ out of $- 4 x (x^{2} + 9)$ .

$(x^{2} + 9) (- 4 x) + (5 x^{2} - 1) (x^{2} + 9) = 0$

Factor $x^{2} + 9$ out of $(5 x^{2} - 1) (x^{2} + 9)$ .

$(x^{2} + 9) (- 4 x) + (x^{2} + 9) (5 x^{2} - 1) = 0$

Factor $x^{2} + 9$ out of $(x^{2} + 9) (- 4 x) + (x^{2} + 9) (5 x^{2} - 1)$ .

$(x^{2} + 9) (- 4 x + 5 x^{2} - 1) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} + 9) (- 4 u + 5 u^{2} - 1) = 0$

Factor by grouping.

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Reorder terms.

$(x^{2} + 9) (5 u^{2} - 4 u - 1) = 0$

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 $a \cdot c = 5 \cdot - 1 = - 5$ and whose sum is $b = - 4$ .

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Factor $- 4$ out of $- 4 u$ .

$(x^{2} + 9) (5 u^{2} - 4 u - 1) = 0$

Rewrite $- 4$ as $1$ plus $- 5$

$(x^{2} + 9) (5 u^{2} + (1 - 5) u - 1) = 0$

Apply the distributive property.

$(x^{2} + 9) (5 u^{2} + 1 u - 5 u - 1) = 0$

Multiply $u$ by $1$ .

$(x^{2} + 9) (5 u^{2} + u - 5 u - 1) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(x^{2} + 9) ((5 u^{2} + u) - 5 u - 1) = 0$

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

$(x^{2} + 9) (u (5 u + 1) - (5 u + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 u + 1$ .

$(x^{2} + 9) ((5 u + 1) (u - 1)) = 0$

Factor.

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Replace all occurrences of $u$ with $x$ .

$(x^{2} + 9) ((5 x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 9) (5 x + 1) (x - 1) = 0$

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

$x^{2} + 9 = 0$

$5 x + 1 = 0$

$x - 1 = 0$

Set $x^{2} + 9$ equal to $0$ and solve for $x$ .

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Set $x^{2} + 9$ equal to $0$ .

$x^{2} + 9 = 0$

Solve $x^{2} + 9 = 0$ for $x$ .

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Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 9}$

Simplify $\pm \sqrt{- 9}$ .

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Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i, - 3 i$

Set $5 x + 1$ equal to $0$ and solve for $x$ .

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Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

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Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

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Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

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Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

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Move the negative in front of the fraction.

$x = - \frac{1}{5}$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x^{2} + 9) (5 x + 1) (x - 1) = 0$ true.

$x = 3 i, - 3 i, - \frac{1}{5}, 1$