Find the Roots/Zeros Using the Rational Roots Test 7x^4-15x^3-33x^2+75x-10

$7 x^{4} - 15 x^{3} - 33 x^{2} + 75 x - 10$

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.

$p = \pm 1, \pm 2, \pm 5, \pm 10$

$q = \pm 1, \pm 7$

Step 2

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

$\pm 1, \pm \frac{1}{7}, \pm 2, \pm \frac{2}{7}, \pm 5, \pm \frac{5}{7}, \pm 10, \pm \frac{10}{7}$

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.

$7 {(2)}^{4} - 15 {(2)}^{3} - 33 {(2)}^{2} + 75 (2) - 10$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$7 \cdot 16 - 15 {(2)}^{3} - 33 {(2)}^{2} + 75 (2) - 10$

Multiply $7$ by $16$ .

$112 - 15 {(2)}^{3} - 33 {(2)}^{2} + 75 (2) - 10$

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

$112 - 15 \cdot 8 - 33 {(2)}^{2} + 75 (2) - 10$

Multiply $- 15$ by $8$ .

$112 - 120 - 33 {(2)}^{2} + 75 (2) - 10$

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

$112 - 120 - 33 \cdot 4 + 75 (2) - 10$

Multiply $- 33$ by $4$ .

$112 - 120 - 132 + 75 (2) - 10$

Multiply $75$ by $2$ .

$112 - 120 - 132 + 150 - 10$

Simplify by adding and subtracting.

Subtract $120$ from $112$ .

$- 8 - 132 + 150 - 10$

Subtract $132$ from $- 8$ .

$- 140 + 150 - 10$

Add $- 140$ and $150$ .

$10 - 10$

Subtract $10$ from $10$ .

$0$

Step 5

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.

$\frac{7 x^{4} - 15 x^{3} - 33 x^{2} + 75 x - 10}{x - 2}$

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.

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0				\xa0

The first number in the dividend $(7)$ is put into the first position of the result area (below the horizontal line).

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0				\xa0
	$7$

Multiply the newest entry in the result $(7)$ by the divisor $(2)$ and place the result of $(14)$ under the next term in the dividend $(- 15)$ .

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$			\xa0
	$7$

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

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$			\xa0
	$7$	$- 1$

Multiply the newest entry in the result $(- 1)$ by the divisor $(2)$ and place the result of $(- 2)$ under the next term in the dividend $(- 33)$ .

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$	$- 2$		\xa0
	$7$	$- 1$

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

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$	$- 2$		\xa0
	$7$	$- 1$	$- 35$

Multiply the newest entry in the result $(- 35)$ by the divisor $(2)$ and place the result of $(- 70)$ under the next term in the dividend $(75)$ .

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$	$- 2$	$- 70$	\xa0
	$7$	$- 1$	$- 35$

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

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$	$- 2$	$- 70$	\xa0
	$7$	$- 1$	$- 35$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(2)$ and place the result of $(10)$ under the next term in the dividend $(- 10)$ .

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$	$- 2$	$- 70$	$10$
	$7$	$- 1$	$- 35$	$5$

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

$2$	$7$	$- 15$	$- 33$	$75$	$- 10$
	\xa0	$14$	$- 2$	$- 70$	$10$
	$7$	$- 1$	$- 35$	$5$	$0$

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

$7 x^{3} + - 1 x^{2} + (- 35) x + 5$

Simplify the quotient polynomial.

$7 x^{3} - x^{2} - 35 x + 5$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) (7 x^{3} - x^{2}) - 35 x + 5$

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

$(x - 2) + x^{2} (7 x - 1) - 5 (7 x - 1)$

$(x - 2) x^{2} (7 x - 1) - 5 (7 x - 1)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $7 x - 1$ .

$(x - 2) (7 x - 1) (x^{2} - 5)$

Step 9

Factor the left side of the equation.

Regroup terms.

$- 15 x^{3} + 75 x + 7 x^{4} - 33 x^{2} - 10 = 0$

Factor $- 15 x$ out of $- 15 x^{3} + 75 x$ .

Factor $- 15 x$ out of $- 15 x^{3}$ .

$- 15 x \cdot x^{2} + 75 x + 7 x^{4} - 33 x^{2} - 10 = 0$

Factor $- 15 x$ out of $75 x$ .

$- 15 x \cdot x^{2} - 15 x \cdot - 5 + 7 x^{4} - 33 x^{2} - 10 = 0$

Factor $- 15 x$ out of $- 15 x (x^{2}) - 15 x (- 5)$ .

$- 15 x (x^{2} - 5) + 7 x^{4} - 33 x^{2} - 10 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 15 x (x^{2} - 5) + 7 {(x^{2})}^{2} - 33 x^{2} - 10 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 15 x (x^{2} - 5) + 7 u^{2} - 33 u - 10 = 0$

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 $a \cdot c = 7 \cdot - 10 = - 70$ and whose sum is $b = - 33$ .

Factor $- 33$ out of $- 33 u$ .

$- 15 x (x^{2} - 5) + 7 u^{2} - 33 u - 10 = 0$

Rewrite $- 33$ as $2$ plus $- 35$

$- 15 x (x^{2} - 5) + 7 u^{2} + (2 - 35) u - 10 = 0$

Apply the distributive property.

$- 15 x (x^{2} - 5) + 7 u^{2} + 2 u - 35 u - 10 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 15 x (x^{2} - 5) + 7 u^{2} + 2 u - 35 u - 10 = 0$

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

$- 15 x (x^{2} - 5) + u (7 u + 2) - 5 (7 u + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 u + 2$ .

$- 15 x (x^{2} - 5) + (7 u + 2) (u - 5) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$- 15 x (x^{2} - 5) + (7 x^{2} + 2) (x^{2} - 5) = 0$

Factor $x^{2} - 5$ out of $- 15 x (x^{2} - 5) + (7 x^{2} + 2) (x^{2} - 5)$ .

Factor $x^{2} - 5$ out of $- 15 x (x^{2} - 5)$ .

$(x^{2} - 5) (- 15 x) + (7 x^{2} + 2) (x^{2} - 5) = 0$

Factor $x^{2} - 5$ out of $(7 x^{2} + 2) (x^{2} - 5)$ .

$(x^{2} - 5) (- 15 x) + (x^{2} - 5) (7 x^{2} + 2) = 0$

Factor $x^{2} - 5$ out of $(x^{2} - 5) (- 15 x) + (x^{2} - 5) (7 x^{2} + 2)$ .

$(x^{2} - 5) (- 15 x + 7 x^{2} + 2) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} - 5) (- 15 u + 7 u^{2} + 2) = 0$

Factor by grouping.

Reorder terms.

$(x^{2} - 5) (7 u^{2} - 15 u + 2) = 0$

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 $a \cdot c = 7 \cdot 2 = 14$ and whose sum is $b = - 15$ .

Factor $- 15$ out of $- 15 u$ .

$(x^{2} - 5) (7 u^{2} - 15 u + 2) = 0$

Rewrite $- 15$ as $- 1$ plus $- 14$

$(x^{2} - 5) (7 u^{2} + (- 1 - 14) u + 2) = 0$

Apply the distributive property.

$(x^{2} - 5) (7 u^{2} - 1 u - 14 u + 2) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{2} - 5) ((7 u^{2} - 1 u) - 14 u + 2) = 0$

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

$(x^{2} - 5) (u (7 u - 1) - 2 (7 u - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 u - 1$ .

$(x^{2} - 5) ((7 u - 1) (u - 2)) = 0$

Factor.

Replace all occurrences of $u$ with $x$ .

$(x^{2} - 5) ((7 x - 1) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x^{2} - 5) (7 x - 1) (x - 2) = 0$

Step 10

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

$x^{2} - 5 = 0$

$7 x - 1 = 0$

$x - 2 = 0$

Step 11

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

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

$x^{2} - 5 = 0$

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

Add $5$ to both sides of the equation.

$x^{2} = 5$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{5}, - \sqrt{5}$

Step 12

Set $7 x - 1$ equal to $0$ and solve for $x$ .

Set $7 x - 1$ equal to $0$ .

$7 x - 1 = 0$

Solve $7 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$7 x = 1$

Divide each term in $7 x = 1$ by $7$ and simplify.

Divide each term in $7 x = 1$ by $7$ .

$\frac{7 x}{7} = \frac{1}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{1}{7}$

Divide $x$ by $1$ .

$x = \frac{1}{7}$

Step 13

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.

$x = 2$

Step 14

The final solution is all the values that make $(x^{2} - 5) (7 x - 1) (x - 2) = 0$ true.

$x = \sqrt{5}, - \sqrt{5}, \frac{1}{7}, 2$

Step 15

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{5}, - \sqrt{5}, \frac{1}{7}, 2$

Decimal Form:

$x = 2.23606797 \dots, - 2.23606797 \dots, 0.\overline{142857}, 2$

Step 16

Find the Roots/Zeros Using the Rational Roots Test 7x^4-15x^3-33x^2+75x-10

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