Find the Roots (Zeros) f(x)=4x^3-13x^2+9x+2

$f (x) = 4 x^{3} - 13 x^{2} + 9 x + 2$

Step 1

Set $4 x^{3} - 13 x^{2} + 9 x + 2$ equal to $0$ .

$4 x^{3} - 13 x^{2} + 9 x + 2 = 0$

Step 2

Solve for $x$ .

Factor $4 x^{3} - 13 x^{2} + 9 x + 2$ 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.

$p = \pm 1, \pm 2 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 2$

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.

$4 \cdot 2^{3} - 13 \cdot 2^{2} + 9 \cdot 2 + 2$

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

$4 \cdot 8 - 13 \cdot 2^{2} + 9 \cdot 2 + 2$

Multiply $4$ by $8$ .

$32 - 13 \cdot 2^{2} + 9 \cdot 2 + 2$

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

$32 - 13 \cdot 4 + 9 \cdot 2 + 2$

Multiply $- 13$ by $4$ .

$32 - 52 + 9 \cdot 2 + 2$

Subtract $52$ from $32$ .

$- 20 + 9 \cdot 2 + 2$

Multiply $9$ by $2$ .

$- 20 + 18 + 2$

Add $- 20$ and $18$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

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.

$\frac{4 x^{3} - 13 x^{2} + 9 x + 2}{x - 2}$

Divide $4 x^{3} - 13 x^{2} + 9 x + 2$ 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 $0$ .

$x$

$2$

$4 x^{3}$

$13 x^{2}$

$9 x$

$2$

Divide the highest order term in the dividend $4 x^{3}$ by the highest order term in divisor $x$ .

					$4 x^{2}$
	$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			+	$4 x^{3}$	-	$8 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{3} - 8 x^{2}$

				$4 x^{2}$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$

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

				$4 x^{2}$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$

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

				$4 x^{2}$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$4 x^{2}$	-	$5 x$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$5 x$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					-	$5 x^{2}$	+	$10 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} + 10 x$

				$4 x^{2}$	-	$5 x$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$

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

				$4 x^{2}$	-	$5 x$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$
							-	$x$

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

				$4 x^{2}$	-	$5 x$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$
							-	$x$	+	$2$

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

				$4 x^{2}$	-	$5 x$	-	$1$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$
							-	$x$	+	$2$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$5 x$	-	$1$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$
							-	$x$	+	$2$
							-	$x$	+	$2$

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

				$4 x^{2}$	-	$5 x$	-	$1$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$
							-	$x$	+	$2$
							+	$x$	-	$2$

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

				$4 x^{2}$	-	$5 x$	-	$1$
$x$	-	$2$		$4 x^{3}$	-	$13 x^{2}$	+	$9 x$	+	$2$
			-	$4 x^{3}$	+	$8 x^{2}$
					-	$5 x^{2}$	+	$9 x$
					+	$5 x^{2}$	-	$10 x$
							-	$x$	+	$2$
							+	$x$	-	$2$
										$0$

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

$4 x^{2} - 5 x - 1$

Write $4 x^{3} - 13 x^{2} + 9 x + 2$ as a set of factors.

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

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

$x - 2 = 0$

$4 x^{2} - 5 x - 1 = 0$

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 $4 x^{2} - 5 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x^{2} - 5 x - 1$ equal to $0$ .

$4 x^{2} - 5 x - 1 = 0$

Solve $4 x^{2} - 5 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 5$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (4 \cdot - 1)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 4 \cdot - 1}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 1$ .

Multiply $- 4$ by $4$ .

$x = \frac{5 \pm \sqrt{25 - 16 \cdot - 1}}{2 \cdot 4}$

Multiply $- 16$ by $- 1$ .

$x = \frac{5 \pm \sqrt{25 + 16}}{2 \cdot 4}$

Add $25$ and $16$ .

$x = \frac{5 \pm \sqrt{41}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{5 \pm \sqrt{41}}{8}$

The final answer is the combination of both solutions.

$x = \frac{5 + \sqrt{41}}{8}, \frac{5 - \sqrt{41}}{8}$

The final solution is all the values that make $(x - 2) (4 x^{2} - 5 x - 1) = 0$ true.

$x = 2, \frac{5 + \sqrt{41}}{8}, \frac{5 - \sqrt{41}}{8}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 2, \frac{5 + \sqrt{41}}{8}, \frac{5 - \sqrt{41}}{8}$

Decimal Form:

$x = 2, 1.42539052 \dots, - 0.17539052 \dots$

Step 4

Find the Roots (Zeros) f(x)=4x^3-13x^2+9x+2

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