Precalculus Examples

$t^{4} - 12 t^{2} + 27$

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.

$p = \pm 1, \pm 3, \pm 9, \pm 27$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 9, \pm 27$

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.

${(3)}^{4} - 12 {(3)}^{2} + 27$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $t = 3$ 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 $3$ to the power of $4$ .

$81 - 12 {(3)}^{2} + 27$

Step 4.1.2

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

$81 - 12 \cdot 9 + 27$

Step 4.1.3

Multiply $- 12$ by $9$ .

$81 - 108 + 27$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $108$ from $81$ .

$- 27 + 27$

Step 4.2.2

Add $- 27$ and $27$ .

$0$

Step 5

Since $3$ is a known root, divide the polynomial by $t - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{t^{4} - 12 t^{2} + 27}{t - 3}$

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.

$3$	$1$	$0$	$- 12$	$0$	$27$

Step 6.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$3$	$1$	$0$	$- 12$	$0$	$27$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(3)$ and place the result of $(3)$ under the next term in the dividend $(0)$ .

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$
	$1$

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.

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$
	$1$	$3$

Step 6.5

Multiply the newest entry in the result $(3)$ by the divisor $(3)$ and place the result of $(9)$ under the next term in the dividend $(- 12)$ .

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$	$9$
	$1$	$3$

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.

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$	$9$
	$1$	$3$	$- 3$

Step 6.7

Multiply the newest entry in the result $(- 3)$ by the divisor $(3)$ and place the result of $(- 9)$ under the next term in the dividend $(0)$ .

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$	$9$	$- 9$
	$1$	$3$	$- 3$

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.

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$	$9$	$- 9$
	$1$	$3$	$- 3$	$- 9$

Step 6.9

Multiply the newest entry in the result $(- 9)$ by the divisor $(3)$ and place the result of $(- 27)$ under the next term in the dividend $(27)$ .

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$	$9$	$- 9$	$- 27$
	$1$	$3$	$- 3$	$- 9$

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.

$3$	$1$	$0$	$- 12$	$0$	$27$
		$3$	$9$	$- 9$	$- 27$
	$1$	$3$	$- 3$	$- 9$	$0$

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.

$1 t^{3} + 3 t^{2} + (- 3) t - 9$

Step 6.12

Simplify the quotient polynomial.

$t^{3} + 3 t^{2} - 3 t - 9$

Step 7

Factor out the greatest common factor from each group.

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Step 7.1

Group the first two terms and the last two terms.

$(t - 3) (t^{3} + 3 t^{2}) - 3 t - 9$

Step 7.2

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

$(t - 3) + t^{2} (t + 3) - 3 (t + 3)$

$(t - 3) t^{2} (t + 3) - 3 (t + 3)$

Step 8

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

$(t - 3) (t + 3) (t^{2} - 3)$

Step 9

Substitute $u = t^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 12 u + 27 = 0$

$u = t^{2}$

Step 10

Factor $u^{2} - 12 u + 27$ using the AC method.

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Step 10.1

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 $27$ and whose sum is $- 12$ .

$- 9, - 3$

Step 10.2

Write the factored form using these integers.

$(u - 9) (u - 3) = 0$

Step 11

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

$u - 9 = 0$

$u - 3 = 0$

Step 12

Set $u - 9$ equal to $0$ and solve for $u$ .

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Step 12.1

Set $u - 9$ equal to $0$ .

$u - 9 = 0$

Step 12.2

Add $9$ to both sides of the equation.

$u = 9$

Step 13

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

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Step 13.1

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

$u - 3 = 0$

Step 13.2

Add $3$ to both sides of the equation.

$u = 3$

Step 14

The final solution is all the values that make $(u - 9) (u - 3) = 0$ true.

$u = 9, 3$

Step 15

Substitute the real value of $u = t^{2}$ back into the solved equation.

$t^{2} = 9$

${(t^{2})}^{1} = 3$

Step 16

Solve the first equation for $t$ .

$t^{2} = 9$

Step 17

Solve the equation for $t$ .

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Step 17.1

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

$t = \pm \sqrt{9}$

Step 17.2

Simplify $\pm \sqrt{9}$ .

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Step 17.2.1

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

$t = \pm \sqrt{3^{2}}$

Step 17.2.2

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

$t = \pm 3$

Step 17.3

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

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Step 17.3.1

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

$t = 3$

Step 17.3.2

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

$t = - 3$

Step 17.3.3

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

$t = 3, - 3$

Step 18

Solve the second equation for $t$ .

${(t^{2})}^{1} = 3$

Step 19

Solve the equation for $t$ .

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Step 19.1

Remove parentheses.

$t^{2} = 3$

Step 19.2

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

$t = \pm \sqrt{3}$

Step 19.3

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

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Step 19.3.1

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

$t = \sqrt{3}$

Step 19.3.2

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

$t = - \sqrt{3}$

Step 19.3.3

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

$t = \sqrt{3}, - \sqrt{3}$

Step 20

The solution to $t^{4} - 12 t^{2} + 27 = 0$ is $t = 3, - 3, \sqrt{3}, - \sqrt{3}$ .

$t = 3, - 3, \sqrt{3}, - \sqrt{3}$

Step 21

The result can be shown in multiple forms.

Exact Form:

$t = 3, - 3, \sqrt{3}, - \sqrt{3}$

Decimal Form:

$t = 3, - 3, 1.73205080 \dots, - 1.73205080 \dots$

Step 22