Find All Complex Solutions 48x^3+40x^2-x-3=0

$48 x^{3} + 40 x^{2} - x - 3 = 0$

Step 1

Factor the left side of the equation.

Factor $48 x^{3} + 40 x^{2} - x - 3$ 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.

$p = \pm 1, \pm 3 q = \pm 1, \pm 48, \pm 2, \pm 24, \pm 3, \pm 16, \pm 4, \pm 12, \pm 6, \pm 8$

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

$\pm 1, \pm 0.0208\overline{3}, \pm 0.5, \pm 0.041\overline{6}, \pm 0.\overline{3}, \pm 0.0625, \pm 0.25, \pm 0.08\overline{3}, \pm 0.1\overline{6}, \pm 0.125, \pm 3, \pm 1.5, \pm 0.1875, \pm 0.75, \pm 0.375$

Substitute $- 0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{3}$ is a root of the polynomial.

Substitute $- 0.\overline{3}$ into the polynomial.

$48 {(- 0.\overline{3})}^{3} + 40 {(- 0.\overline{3})}^{2} + 0.\overline{3} - 3$

Raise $- 0.\overline{3}$ to the power of $3$ .

$48 \cdot - 0.\overline{037} + 40 {(- 0.\overline{3})}^{2} + 0.\overline{3} - 3$

Multiply $48$ by $- 0.\overline{037}$ .

$- 1.\overline{7} + 40 {(- 0.\overline{3})}^{2} + 0.\overline{3} - 3$

Raise $- 0.\overline{3}$ to the power of $2$ .

$- 1.\overline{7} + 40 \cdot 0.\overline{1} + 0.\overline{3} - 3$

Multiply $40$ by $0.\overline{1}$ .

$- 1.\overline{7} + 4.\overline{4} + 0.\overline{3} - 3$

Add $- 1.\overline{7}$ and $4.\overline{4}$ .

$2.\overline{6} + 0.\overline{3} - 3$

Add $2.\overline{6}$ and $0.\overline{3}$ .

$3 - 3$

Subtract $3$ from $3$ .

$0$

Since $- 0.\overline{3}$ is a known root, divide the polynomial by $3 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{48 x^{3} + 40 x^{2} - x - 3}{3 x + 1}$

Divide $48 x^{3} + 40 x^{2} - x - 3$ by $3 x + 1$ .

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

$3 x$

$1$

$48 x^{3}$

$40 x^{2}$

$x$

$3$

Divide the highest order term in the dividend $48 x^{3}$ by the highest order term in divisor $3 x$ .

					$16 x^{2}$
	$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$

Multiply the new quotient term by the divisor.

				$16 x^{2}$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			+	$48 x^{3}$	+	$16 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $48 x^{3} + 16 x^{2}$

				$16 x^{2}$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$

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

				$16 x^{2}$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$

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

				$16 x^{2}$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$

Divide the highest order term in the dividend $24 x^{2}$ by the highest order term in divisor $3 x$ .

				$16 x^{2}$	+	$8 x$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$

Multiply the new quotient term by the divisor.

				$16 x^{2}$	+	$8 x$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					+	$24 x^{2}$	+	$8 x$

The expression needs to be subtracted from the dividend, so change all the signs in $24 x^{2} + 8 x$

				$16 x^{2}$	+	$8 x$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$

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

				$16 x^{2}$	+	$8 x$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$
							-	$9 x$

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

				$16 x^{2}$	+	$8 x$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$
							-	$9 x$	-	$3$

Divide the highest order term in the dividend $- 9 x$ by the highest order term in divisor $3 x$ .

				$16 x^{2}$	+	$8 x$	-	$3$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$
							-	$9 x$	-	$3$

Multiply the new quotient term by the divisor.

				$16 x^{2}$	+	$8 x$	-	$3$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$
							-	$9 x$	-	$3$
							-	$9 x$	-	$3$

The expression needs to be subtracted from the dividend, so change all the signs in $- 9 x - 3$

				$16 x^{2}$	+	$8 x$	-	$3$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$
							-	$9 x$	-	$3$
							+	$9 x$	+	$3$

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

				$16 x^{2}$	+	$8 x$	-	$3$
$3 x$	+	$1$		$48 x^{3}$	+	$40 x^{2}$	-	$x$	-	$3$
			-	$48 x^{3}$	-	$16 x^{2}$
					+	$24 x^{2}$	-	$x$
					-	$24 x^{2}$	-	$8 x$
							-	$9 x$	-	$3$
							+	$9 x$	+	$3$
										$0$

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

$16 x^{2} + 8 x - 3$

Write $48 x^{3} + 40 x^{2} - x - 3$ as a set of factors.

$(3 x + 1) (16 x^{2} + 8 x - 3) = 0$

Factor by grouping.

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 $a \cdot c = 16 \cdot - 3 = - 48$ and whose sum is $b = 8$ .

Factor $8$ out of $8 x$ .

$(3 x + 1) (16 x^{2} + 8 (x) - 3) = 0$

Rewrite $8$ as $- 4$ plus $12$

$(3 x + 1) (16 x^{2} + (- 4 + 12) x - 3) = 0$

Apply the distributive property.

$(3 x + 1) (16 x^{2} - 4 x + 12 x - 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x + 1) ((16 x^{2} - 4 x) + 12 x - 3) = 0$

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

$(3 x + 1) (4 x (4 x - 1) + 3 (4 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(3 x + 1) ((4 x - 1) (4 x + 3)) = 0$

Remove unnecessary parentheses.

$(3 x + 1) (4 x - 1) (4 x + 3) = 0$

Step 2

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

$3 x + 1 = 0$

$4 x - 1 = 0$

$4 x + 3 = 0$

Step 3

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Step 4

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 5

Set $4 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Solve $4 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Divide each term in $4 x = - 3$ by $4$ and simplify.

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

Step 6

The final solution is all the values that make $(3 x + 1) (4 x - 1) (4 x + 3) = 0$ true.

$x = - \frac{1}{3}, \frac{1}{4}, - \frac{3}{4}$

Find All Complex Solutions 48x^3+40x^2-x-3=0

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