Find the Roots/Zeros Using the Rational Roots Test 4x^5-11x^3+3x^2+7x-3

$4 x^{5} - 11 x^{3} + 3 x^{2} + 7 x - 3$

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.

$p = \pm 1, \pm 3$

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

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}$

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.

$4 {(1)}^{5} - 11 {(1)}^{3} + 3 {(1)}^{2} + 7 (1) - 3$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

Simplify each term.

One to any power is one.

$4 \cdot 1 - 11 {(1)}^{3} + 3 {(1)}^{2} + 7 (1) - 3$

Multiply $4$ by $1$ .

$4 - 11 {(1)}^{3} + 3 {(1)}^{2} + 7 (1) - 3$

One to any power is one.

$4 - 11 \cdot 1 + 3 {(1)}^{2} + 7 (1) - 3$

Multiply $- 11$ by $1$ .

$4 - 11 + 3 {(1)}^{2} + 7 (1) - 3$

One to any power is one.

$4 - 11 + 3 \cdot 1 + 7 (1) - 3$

Multiply $3$ by $1$ .

$4 - 11 + 3 + 7 (1) - 3$

Multiply $7$ by $1$ .

$4 - 11 + 3 + 7 - 3$

$4 - 11 + 3 + 7 - 3$

Simplify by adding and subtracting.

Subtract $11$ from $4$ .

$- 7 + 3 + 7 - 3$

Add $- 7$ and $3$ .

$- 4 + 7 - 3$

Add $- 4$ and $7$ .

$3 - 3$

Subtract $3$ from $3$ .

$0$

$0$

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{4 x^{5} - 11 x^{3} + 3 x^{2} + 7 x - 3}{x - 1}$

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.

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0					\xa0

The first number in the dividend $(4)$ is put into the first position of the result area (below the horizontal line).

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0					\xa0
	$4$

Multiply the newest entry in the result $(4)$ by the divisor $(1)$ and place the result of $(4)$ under the next term in the dividend $(0)$ .

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$				\xa0
	$4$

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

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$				\xa0
	$4$	$4$

Multiply the newest entry in the result $(4)$ by the divisor $(1)$ and place the result of $(4)$ under the next term in the dividend $(- 11)$ .

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$			\xa0
	$4$	$4$

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

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$			\xa0
	$4$	$4$	$- 7$

Multiply the newest entry in the result $(- 7)$ by the divisor $(1)$ and place the result of $(- 7)$ under the next term in the dividend $(3)$ .

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$	$- 7$		\xa0
	$4$	$4$	$- 7$

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

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$	$- 7$		\xa0
	$4$	$4$	$- 7$	$- 4$

Multiply the newest entry in the result $(- 4)$ by the divisor $(1)$ and place the result of $(- 4)$ under the next term in the dividend $(7)$ .

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$	$- 7$	$- 4$	\xa0
	$4$	$4$	$- 7$	$- 4$

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

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$	$- 7$	$- 4$	\xa0
	$4$	$4$	$- 7$	$- 4$	$3$

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

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$	$- 7$	$- 4$	$3$
	$4$	$4$	$- 7$	$- 4$	$3$

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

$1$	$4$	$0$	$- 11$	$3$	$7$	$- 3$
	\xa0	$4$	$4$	$- 7$	$- 4$	$3$
	$4$	$4$	$- 7$	$- 4$	$3$	$0$

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

$4 x^{4} + 4 x^{3} - 7 x^{2} + - 4 x + 3$

Simplify the quotient polynomial.

$4 x^{4} + 4 x^{3} - 7 x^{2} - 4 x + 3$

$4 x^{4} + 4 x^{3} - 7 x^{2} - 4 x + 3$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Regroup terms.

$4 x^{4} + 4 x^{3} - 7 x^{2} - 4 x + 3 = 0$

Factor $4 x^{3}$ out of $4 x^{4} + 4 x^{3}$ .

Factor $4 x^{3}$ out of $4 x^{4}$ .

$4 x^{3} (x) + 4 x^{3} - 7 x^{2} - 4 x + 3 = 0$

Factor $4 x^{3}$ out of $4 x^{3}$ .

$4 x^{3} (x) + 4 x^{3} (1) - 7 x^{2} - 4 x + 3 = 0$

Factor $4 x^{3}$ out of $4 x^{3} (x) + 4 x^{3} (1)$ .

$4 x^{3} (x + 1) - 7 x^{2} - 4 x + 3 = 0$

$4 x^{3} (x + 1) - 7 x^{2} - 4 x + 3 = 0$

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 $a \cdot c = - 7 \cdot 3 = - 21$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 x$ .

$4 x^{3} (x + 1) - 7 x^{2} - 4 x + 3 = 0$

Rewrite $- 4$ as $3$ plus $- 7$

$4 x^{3} (x + 1) - 7 x^{2} + (3 - 7) x + 3 = 0$

Apply the distributive property.

$4 x^{3} (x + 1) - 7 x^{2} + 3 x - 7 x + 3 = 0$

$4 x^{3} (x + 1) - 7 x^{2} + 3 x - 7 x + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$4 x^{3} (x + 1) - 7 x^{2} + 3 x - 7 x + 3 = 0$

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

$4 x^{3} (x + 1) + x (- 7 x + 3) + 1 (- 7 x + 3) = 0$

$4 x^{3} (x + 1) + x (- 7 x + 3) + 1 (- 7 x + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $- 7 x + 3$ .

$4 x^{3} (x + 1) + (- 7 x + 3) (x + 1) = 0$

$4 x^{3} (x + 1) + (- 7 x + 3) (x + 1) = 0$

Factor $x + 1$ out of $4 x^{3} (x + 1) + (- 7 x + 3) (x + 1)$ .

Factor $x + 1$ out of $4 x^{3} (x + 1)$ .

$(x + 1) (4 x^{3}) + (- 7 x + 3) (x + 1) = 0$

Factor $x + 1$ out of $(- 7 x + 3) (x + 1)$ .

$(x + 1) (4 x^{3}) + (x + 1) (- 7 x + 3) = 0$

Factor $x + 1$ out of $(x + 1) (4 x^{3}) + (x + 1) (- 7 x + 3)$ .

$(x + 1) (4 x^{3} - 7 x + 3) = 0$

$(x + 1) (4 x^{3} - 7 x + 3) = 0$

Factor.

Rewrite $4 x^{3} - 7 x + 3$ in a factored form.

Factor $4 x^{3} - 7 x + 3$ 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.

$p = \pm 1, \pm 3 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 3, \pm 0.75, \pm 1.5$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$4 \cdot 1^{3} - 7 \cdot 1 + 3$

Raise $1$ to the power of $3$ .

$4 \cdot 1 - 7 \cdot 1 + 3$

Multiply $4$ by $1$ .

$4 - 7 \cdot 1 + 3$

Multiply $- 7$ by $1$ .

$4 - 7 + 3$

Subtract $7$ from $4$ .

$- 3 + 3$

Add $- 3$ and $3$ .

$0$

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{4 x^{3} - 7 x + 3}{x - 1}$

Divide $4 x^{3} - 7 x + 3$ by $x - 1$ .

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

$x$

-

$1$

$4 x^{3}$

+

$0 x^{2}$

-

$7 x$

+

$3$

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

					$4 x^{2}$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$

Multiply the new quotient term by the divisor.

					$4 x^{2}$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				+	$4 x^{3}$	-	$4 x^{2}$

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

					$4 x^{2}$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$

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

					$4 x^{2}$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$

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

					$4 x^{2}$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$

Divide the highest order term in the dividend $4 x^{2}$ by the highest order term in divisor $x$ .

					$4 x^{2}$	+	$4 x$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$

Multiply the new quotient term by the divisor.

					$4 x^{2}$	+	$4 x$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						+	$4 x^{2}$	-	$4 x$

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

					$4 x^{2}$	+	$4 x$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$

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

					$4 x^{2}$	+	$4 x$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$
								-	$3 x$

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

					$4 x^{2}$	+	$4 x$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$
								-	$3 x$	+	$3$

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

					$4 x^{2}$	+	$4 x$	-	$3$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$
								-	$3 x$	+	$3$

Multiply the new quotient term by the divisor.

					$4 x^{2}$	+	$4 x$	-	$3$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$
								-	$3 x$	+	$3$
								-	$3 x$	+	$3$

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

					$4 x^{2}$	+	$4 x$	-	$3$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$
								-	$3 x$	+	$3$
								+	$3 x$	-	$3$

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

					$4 x^{2}$	+	$4 x$	-	$3$
	$x$	-	$1$		$4 x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$3$
				-	$4 x^{3}$	+	$4 x^{2}$
						+	$4 x^{2}$	-	$7 x$
						-	$4 x^{2}$	+	$4 x$
								-	$3 x$	+	$3$
								+	$3 x$	-	$3$
											$0$

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

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

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

Write $4 x^{3} - 7 x + 3$ as a set of factors.

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

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

Factor by grouping.

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 $a \cdot c = 4 \cdot - 3 = - 12$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

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

Rewrite $4$ as $- 2$ plus $6$

$(x + 1) ((x - 1) (4 x^{2} + (- 2 + 6) x - 3)) = 0$

Apply the distributive property.

$(x + 1) ((x - 1) (4 x^{2} - 2 x + 6 x - 3)) = 0$

$(x + 1) ((x - 1) (4 x^{2} - 2 x + 6 x - 3)) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((x - 1) ((4 x^{2} - 2 x) + 6 x - 3)) = 0$

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

$(x + 1) ((x - 1) (2 x (2 x - 1) + 3 (2 x - 1))) = 0$

$(x + 1) ((x - 1) (2 x (2 x - 1) + 3 (2 x - 1))) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(x + 1) ((x - 1) ((2 x - 1) (2 x + 3))) = 0$

$(x + 1) ((x - 1) ((2 x - 1) (2 x + 3))) = 0$

Remove unnecessary parentheses.

$(x + 1) ((x - 1) (2 x - 1) (2 x + 3)) = 0$

$(x + 1) ((x - 1) (2 x - 1) (2 x + 3)) = 0$

$(x + 1) ((x - 1) (2 x - 1) (2 x + 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 1) (2 x - 1) (2 x + 3) = 0$

$(x + 1) (x - 1) (2 x - 1) (2 x + 3) = 0$

$(x + 1) (x - 1) (2 x - 1) (2 x + 3) = 0$

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

$x + 1 = 0$

$x - 1 = 0$

$2 x - 1 = 0$

$2 x + 3 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

$x = - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

$x = 1$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

$x = \frac{1}{2}$

$x = \frac{1}{2}$

$x = \frac{1}{2}$

$x = \frac{1}{2}$

$x = \frac{1}{2}$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

$x = \frac{- 3}{2}$

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

$x = - \frac{3}{2}$

$x = - \frac{3}{2}$

$x = - \frac{3}{2}$

$x = - \frac{3}{2}$

The final solution is all the values that make $(x + 1) (x - 1) (2 x - 1) (2 x + 3) = 0$ true.

$x = - 1, 1, \frac{1}{2}, - \frac{3}{2}$

$x = - 1, 1, \frac{1}{2}, - \frac{3}{2}$

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

$(x - 1) (x + 1) (x - \frac{1}{2}) (x + \frac{3}{2})$

These are the roots (zeros) of the polynomial $4 x^{5} - 11 x^{3} + 3 x^{2} + 7 x - 3$ .

$x = 1, - 1, \frac{1}{2}, - \frac{3}{2}$

Related Questions