Find the Roots/Zeros Using the Rational Roots Test f(x)=4x^3+7x^2-62x+15

$f (x) = 4 x^{3} + 7 x^{2} - 62 x + 15$

Step 1

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm 15, \pm \frac{15}{2}, \pm \frac{15}{4}$

Step 3

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.

$4 {(\frac{1}{4})}^{3} + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{4}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{4}$ .

$4 \frac{1^{3}}{4^{3}} + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

One to any power is one.

$4 \frac{1}{4^{3}} + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

Raise $4$ to the power of $3$ .

$4 (\frac{1}{64}) + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

Cancel the common factor of $4$ .

Factor $4$ out of $64$ .

$4 \frac{1}{4 (16)} + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

Cancel the common factor.

$4 \frac{1}{4 \cdot 16} + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

Rewrite the expression.

$\frac{1}{16} + 7 {(\frac{1}{4})}^{2} - 62 (\frac{1}{4}) + 15$

Apply the product rule to $\frac{1}{4}$ .

$\frac{1}{16} + 7 \frac{1^{2}}{4^{2}} - 62 (\frac{1}{4}) + 15$

One to any power is one.

$\frac{1}{16} + 7 \frac{1}{4^{2}} - 62 (\frac{1}{4}) + 15$

Raise $4$ to the power of $2$ .

$\frac{1}{16} + 7 (\frac{1}{16}) - 62 (\frac{1}{4}) + 15$

Combine $7$ and $\frac{1}{16}$ .

$\frac{1}{16} + \frac{7}{16} - 62 (\frac{1}{4}) + 15$

Cancel the common factor of $2$ .

Factor $2$ out of $- 62$ .

$\frac{1}{16} + \frac{7}{16} + 2 (- 31) \frac{1}{4} + 15$

Factor $2$ out of $4$ .

$\frac{1}{16} + \frac{7}{16} + 2 \cdot - 31 \frac{1}{2 \cdot 2} + 15$

Cancel the common factor.

$\frac{1}{16} + \frac{7}{16} + 2 \cdot - 31 \frac{1}{2 \cdot 2} + 15$

Rewrite the expression.

$\frac{1}{16} + \frac{7}{16} - 31 (\frac{1}{2}) + 15$

Combine $- 31$ and $\frac{1}{2}$ .

$\frac{1}{16} + \frac{7}{16} + \frac{- 31}{2} + 15$

Move the negative in front of the fraction.

$\frac{1}{16} + \frac{7}{16} - \frac{31}{2} + 15$

Combine fractions.

Combine the numerators over the common denominator.

$15 + \frac{1 + 7}{16} + \frac{- 31}{2}$

Add $1$ and $7$ .

$15 + \frac{8}{16} + \frac{- 31}{2}$

Find the common denominator.

Write $15$ as a fraction with denominator $1$ .

$\frac{15}{1} + \frac{8}{16} + \frac{- 31}{2}$

Multiply $\frac{15}{1}$ by $\frac{16}{16}$ .

$\frac{15}{1} \cdot \frac{16}{16} + \frac{8}{16} + \frac{- 31}{2}$

Multiply $\frac{15}{1}$ by $\frac{16}{16}$ .

$\frac{15 \cdot 16}{16} + \frac{8}{16} + \frac{- 31}{2}$

Multiply $\frac{- 31}{2}$ by $\frac{8}{8}$ .

$\frac{15 \cdot 16}{16} + \frac{8}{16} + \frac{- 31}{2} \cdot \frac{8}{8}$

Multiply $\frac{- 31}{2}$ by $\frac{8}{8}$ .

$\frac{15 \cdot 16}{16} + \frac{8}{16} + \frac{- 31 \cdot 8}{2 \cdot 8}$

Multiply $2$ by $8$ .

$\frac{15 \cdot 16}{16} + \frac{8}{16} + \frac{- 31 \cdot 8}{16}$

Combine the numerators over the common denominator.

$\frac{15 \cdot 16 + 8 - 31 \cdot 8}{16}$

Simplify each term.

Multiply $15$ by $16$ .

$\frac{240 + 8 - 31 \cdot 8}{16}$

Multiply $- 31$ by $8$ .

$\frac{240 + 8 - 248}{16}$

Simplify the expression.

Add $240$ and $8$ .

$\frac{248 - 248}{16}$

Subtract $248$ from $248$ .

$\frac{0}{16}$

Divide $0$ by $16$ .

$0$

Step 5

Since $\frac{1}{4}$ is a known root, divide the polynomial by $x - \frac{1}{4}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{4 x^{3} + 7 x^{2} - 62 x + 15}{x - \frac{1}{4}}$

Step 6

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.

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0			\xa0

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

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0			\xa0
	$4$

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

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0	$1$		\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.

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0	$1$		\xa0
	$4$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(\frac{1}{4})$ and place the result of $(2)$ under the next term in the dividend $(- 62)$ .

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0	$1$	$2$	\xa0
	$4$	$8$

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

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0	$1$	$2$	\xa0
	$4$	$8$	$- 60$

Multiply the newest entry in the result $(- 60)$ by the divisor $(\frac{1}{4})$ and place the result of $(- 15)$ under the next term in the dividend $(15)$ .

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0	$1$	$2$	$- 15$
	$4$	$8$	$- 60$

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

$\frac{1}{4}$	$4$	$7$	$- 62$	$15$
	\xa0	$1$	$2$	$- 15$
	$4$	$8$	$- 60$	$0$

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

$4 x^{2} + 8 x - 60$

Step 7

Factor $4$ out of $4 x^{2} + 8 x - 60$ .

Factor $4$ out of $4 x^{2}$ .

$(x - \frac{1}{4}) (4 (x^{2}) + 8 x - 60)$

Factor $4$ out of $8 x$ .

$(x - \frac{1}{4}) (4 (x^{2}) + 4 (2 x) - 60)$

Factor $4$ out of $- 60$ .

$(x - \frac{1}{4}) (4 x^{2} + 4 (2 x) + 4 \cdot - 15)$

Factor $4$ out of $4 x^{2} + 4 (2 x)$ .

$(x - \frac{1}{4}) (4 (x^{2} + 2 x) + 4 \cdot - 15)$

Factor $4$ out of $4 (x^{2} + 2 x) + 4 \cdot - 15$ .

$(x - \frac{1}{4}) (4 (x^{2} + 2 x - 15))$

Step 8

Factor.

Factor $x^{2} + 2 x - 15$ 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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$(x - \frac{1}{4}) (4 ((x - 3) (x + 5)))$

Remove unnecessary parentheses.

$(x - \frac{1}{4}) (4 (x - 3) (x + 5))$

Step 9

Factor the left side of the equation.

Factor $4 x^{3} + 7 x^{2} - 62 x + 15$ 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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 15, \pm 3.75, \pm 7.5, \pm 3, \pm 0.75, \pm 1.5, \pm 5, \pm 1.25, \pm 2.5$

Substitute $0.25$ and simplify the expression. In this case, the expression is equal to $0$ so $0.25$ is a root of the polynomial.

Substitute $0.25$ into the polynomial.

$4 \cdot {0.25}^{3} + 7 \cdot {0.25}^{2} - 62 \cdot 0.25 + 15$

Raise $0.25$ to the power of $3$ .

$4 \cdot 0.015625 + 7 \cdot {0.25}^{2} - 62 \cdot 0.25 + 15$

Multiply $4$ by $0.015625$ .

$0.0625 + 7 \cdot {0.25}^{2} - 62 \cdot 0.25 + 15$

Raise $0.25$ to the power of $2$ .

$0.0625 + 7 \cdot 0.0625 - 62 \cdot 0.25 + 15$

Multiply $7$ by $0.0625$ .

$0.0625 + 0.4375 - 62 \cdot 0.25 + 15$

Add $0.0625$ and $0.4375$ .

$0.5 - 62 \cdot 0.25 + 15$

Multiply $- 62$ by $0.25$ .

$0.5 - 15.5 + 15$

Subtract $15.5$ from $0.5$ .

$- 15 + 15$

Add $- 15$ and $15$ .

$0$

Since $0.25$ is a known root, divide the polynomial by $4 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{4 x^{3} + 7 x^{2} - 62 x + 15}{4 x - 1}$

Divide $4 x^{3} + 7 x^{2} - 62 x + 15$ by $4 x - 1$ .

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

$4 x$

$1$

$4 x^{3}$

$7 x^{2}$

$62 x$

$15$

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

					$x^{2}$
	$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$

Multiply the new quotient term by the divisor.

				$x^{2}$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			+	$4 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$

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

				$x^{2}$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$

Divide the highest order term in the dividend $8 x^{2}$ by the highest order term in divisor $4 x$ .

				$x^{2}$	+	$2 x$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					+	$8 x^{2}$	-	$2 x$

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

				$x^{2}$	+	$2 x$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$

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

				$x^{2}$	+	$2 x$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$
							-	$60 x$

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

				$x^{2}$	+	$2 x$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$
							-	$60 x$	+	$15$

Divide the highest order term in the dividend $- 60 x$ by the highest order term in divisor $4 x$ .

				$x^{2}$	+	$2 x$	-	$15$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$
							-	$60 x$	+	$15$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	-	$15$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$
							-	$60 x$	+	$15$
							-	$60 x$	+	$15$

The expression needs to be subtracted from the dividend, so change all the signs in $- 60 x + 15$

				$x^{2}$	+	$2 x$	-	$15$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$
							-	$60 x$	+	$15$
							+	$60 x$	-	$15$

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

				$x^{2}$	+	$2 x$	-	$15$
$4 x$	-	$1$		$4 x^{3}$	+	$7 x^{2}$	-	$62 x$	+	$15$
			-	$4 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$62 x$
					-	$8 x^{2}$	+	$2 x$
							-	$60 x$	+	$15$
							+	$60 x$	-	$15$
										$0$

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

$x^{2} + 2 x - 15$

Write $4 x^{3} + 7 x^{2} - 62 x + 15$ as a set of factors.

$(4 x - 1) (x^{2} + 2 x - 15) = 0$

Factor $x^{2} + 2 x - 15$ 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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$(4 x - 1) ((x - 3) (x + 5)) = 0$

Remove unnecessary parentheses.

$(4 x - 1) (x - 3) (x + 5) = 0$

Step 10

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

$4 x - 1 = 0$

$x - 3 = 0$

$x + 5 = 0$

Step 11

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 12

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.

$x = 3$

Step 13

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 14

The final solution is all the values that make $(4 x - 1) (x - 3) (x + 5) = 0$ true.

$x = \frac{1}{4}, 3, - 5$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=4x^3+7x^2-62x+15

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