Find the Roots (Zeros) f(x)=x^3-9x^2+26x-24

$f (x) = x^{3} - 9 x^{2} + 26 x - 24$

Set $x^{3} - 9 x^{2} + 26 x - 24$ equal to $0$ .

$x^{3} - 9 x^{2} + 26 x - 24 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x^{3} - 9 x^{2} + 26 x - 24$ using the rational roots test.

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.

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

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

$\pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6$

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$2^{3} - 9 \cdot 2^{2} + 26 \cdot 2 - 24$

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

$8 - 9 \cdot 2^{2} + 26 \cdot 2 - 24$

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

$8 - 9 \cdot 4 + 26 \cdot 2 - 24$

Multiply $- 9$ by $4$ .

$8 - 36 + 26 \cdot 2 - 24$

Subtract $36$ from $8$ .

$- 28 + 26 \cdot 2 - 24$

Multiply $26$ by $2$ .

$- 28 + 52 - 24$

Add $- 28$ and $52$ .

$24 - 24$

Subtract $24$ from $24$ .

$0$

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 9 x^{2} + 26 x - 24}{x - 2}$

Divide $x^{3} - 9 x^{2} + 26 x - 24$ by $x - 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x^{3}$

$9 x^{2}$

$26 x$

$24$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			+	$x^{3}$	-	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 2 x^{2}$

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					-	$7 x^{2}$	+	$14 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} + 14 x$

				$x^{2}$	-	$7 x$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$7 x$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$
							+	$12 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$7 x$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$
							+	$12 x$	-	$24$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$	+	$12$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$
							+	$12 x$	-	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$	+	$12$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$
							+	$12 x$	-	$24$
							+	$12 x$	-	$24$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x - 24$

				$x^{2}$	-	$7 x$	+	$12$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$
							+	$12 x$	-	$24$
							-	$12 x$	+	$24$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$7 x$	+	$12$
$x$	-	$2$		$x^{3}$	-	$9 x^{2}$	+	$26 x$	-	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$7 x^{2}$	+	$26 x$
					+	$7 x^{2}$	-	$14 x$
							+	$12 x$	-	$24$
							-	$12 x$	+	$24$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 7 x + 12$

Write $x^{3} - 9 x^{2} + 26 x - 24$ as a set of factors.

$(x - 2) (x^{2} - 7 x + 12) = 0$

Factor $x^{2} - 7 x + 12$ using the AC method.

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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$(x - 2) ((x - 4) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x - 2) (x - 4) (x - 3) = 0$

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

$x - 2 = 0$

$x - 4 = 0$

$x - 3 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x - 2) (x - 4) (x - 3) = 0$ true.

$x = 2, 4, 3$

Find the Roots (Zeros) f(x)=x^3-9x^2+26x-24

Related Questions

Support LCMGCF.com by disabling your adblocker.