Find the Roots (Zeros) t^3-13t^2+65t-105=0

$t^{3} - 13 t^{2} + 65 t - 105 = 0$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 7, \pm 15, \pm 21, \pm 35, \pm 105$

$q = \pm 1$

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

$\pm 1, \pm 3, \pm 5, \pm 7, \pm 15, \pm 21, \pm 35, \pm 105$

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.

${(3)}^{3} - 13 {(3)}^{2} + 65 (3) - 105$

Simplify the expression. In this case, the expression is equal to $0$ so $t = 3$ is a root of the polynomial.

Simplify each term.

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

$27 - 13 {(3)}^{2} + 65 (3) - 105$

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

$27 - 13 \cdot 9 + 65 (3) - 105$

Multiply $- 13$ by $9$ .

$27 - 117 + 65 (3) - 105$

Multiply $65$ by $3$ .

$27 - 117 + 195 - 105$

Simplify by adding and subtracting.

Subtract $117$ from $27$ .

$- 90 + 195 - 105$

Add $- 90$ and $195$ .

$105 - 105$

Subtract $105$ from $105$ .

$0$

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.

$\frac{t^{3} - 13 t^{2} + 65 t - 105}{t - 3}$

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

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

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0			\xa0

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

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0			\xa0
	$1$

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 $(- 13)$ .

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0	$3$		\xa0
	$1$

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

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0	$3$		\xa0
	$1$	$- 10$

Multiply the newest entry in the result $(- 10)$ by the divisor $(3)$ and place the result of $(- 30)$ under the next term in the dividend $(65)$ .

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0	$3$	$- 30$	\xa0
	$1$	$- 10$

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

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0	$3$	$- 30$	\xa0
	$1$	$- 10$	$35$

Multiply the newest entry in the result $(35)$ by the divisor $(3)$ and place the result of $(105)$ under the next term in the dividend $(- 105)$ .

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0	$3$	$- 30$	$105$
	$1$	$- 10$	$35$

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

$3$	$1$	$- 13$	$65$	$- 105$
	\xa0	$3$	$- 30$	$105$
	$1$	$- 10$	$35$	$0$

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

$1 t^{2} + - 10 t + 35$

Simplify the quotient polynomial.

$t^{2} - 10 t + 35$

Solve the equation to find any remaining roots.

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 10$ , and $c = 35$ into the quadratic formula and solve for $t$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot 35)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$t = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 35}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$t = \frac{10 \pm \sqrt{100 - 4 \cdot 35}}{2 \cdot 1}$

Multiply $- 4$ by $35$ .

$t = \frac{10 \pm \sqrt{100 - 140}}{2 \cdot 1}$

Subtract $140$ from $100$ .

$t = \frac{10 \pm \sqrt{- 40}}{2 \cdot 1}$

Rewrite $- 40$ as $- 1 (40)$ .

$t = \frac{10 \pm \sqrt{- 1 \cdot 40}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (40)}$ as $\sqrt{- 1} \cdot \sqrt{40}$ .

$t = \frac{10 \pm \sqrt{- 1} \cdot \sqrt{40}}{2 \cdot 1}$

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

$t = \frac{10 \pm i \cdot \sqrt{40}}{2 \cdot 1}$

Rewrite $40$ as $2^{2} \cdot 10$ .

Factor $4$ out of $40$ .

$t = \frac{10 \pm i \cdot \sqrt{4 (10)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$t = \frac{10 \pm i \cdot \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Pull terms out from under the radical.

$t = \frac{10 \pm i \cdot (2 \sqrt{10})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$t = \frac{10 \pm 2 i \sqrt{10}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{10 \pm 2 i \sqrt{10}}{2}$

Simplify $\frac{10 \pm 2 i \sqrt{10}}{2}$ .

$t = 5 \pm i \sqrt{10}$

The final answer is the combination of both solutions.

$t = 5 + i \sqrt{10}, 5 - i \sqrt{10}$

The polynomial can be written as a set of linear factors.

$(t - 3) (t - 5 - i \sqrt{10}) (t - 5 + \sqrt{10} i)$

These are the roots (zeros) of the polynomial $t^{3} - 13 t^{2} + 65 t - 105$ .

$t = 3, 5 + i \sqrt{10}, 5 - i \sqrt{10}$

Find the Roots (Zeros) t^3-13t^2+65t-105=0

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