Find the Roots/Zeros Using the Rational Roots Test 25x^4+26x^3+51x^2+52x+2

$25 x^{4} + 26 x^{3} + 51 x^{2} + 52 x + 2$

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 5, \pm 25$

Step 2

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

$\pm 1, \pm \frac{1}{5}, \pm \frac{1}{25}, \pm 2, \pm \frac{2}{5}, \pm \frac{2}{25}$

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.

$25 {(- 1)}^{4} + 26 {(- 1)}^{3} + 51 {(- 1)}^{2} + 52 (- 1) + 2$

Step 4

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

Simplify each term.

Raise $- 1$ to the power of $4$ .

$25 \cdot 1 + 26 {(- 1)}^{3} + 51 {(- 1)}^{2} + 52 (- 1) + 2$

Multiply $25$ by $1$ .

$25 + 26 {(- 1)}^{3} + 51 {(- 1)}^{2} + 52 (- 1) + 2$

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

$25 + 26 \cdot - 1 + 51 {(- 1)}^{2} + 52 (- 1) + 2$

Multiply $26$ by $- 1$ .

$25 - 26 + 51 {(- 1)}^{2} + 52 (- 1) + 2$

Raise $- 1$ to the power of $2$ .

$25 - 26 + 51 \cdot 1 + 52 (- 1) + 2$

Multiply $51$ by $1$ .

$25 - 26 + 51 + 52 (- 1) + 2$

Multiply $52$ by $- 1$ .

$25 - 26 + 51 - 52 + 2$

Simplify by adding and subtracting.

Subtract $26$ from $25$ .

$- 1 + 51 - 52 + 2$

Add $- 1$ and $51$ .

$50 - 52 + 2$

Subtract $52$ from $50$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

Step 5

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.

$\frac{25 x^{4} + 26 x^{3} + 51 x^{2} + 52 x + 2}{x + 1}$

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.

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0				\xa0

The first number in the dividend $(25)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0				\xa0
	$25$

Multiply the newest entry in the result $(25)$ by the divisor $(- 1)$ and place the result of $(- 25)$ under the next term in the dividend $(26)$ .

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$			\xa0
	$25$

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

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$			\xa0
	$25$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(51)$ .

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$	$- 1$		\xa0
	$25$	$1$

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

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$	$- 1$		\xa0
	$25$	$1$	$50$

Multiply the newest entry in the result $(50)$ by the divisor $(- 1)$ and place the result of $(- 50)$ under the next term in the dividend $(52)$ .

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$	$- 1$	$- 50$	\xa0
	$25$	$1$	$50$

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

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$	$- 1$	$- 50$	\xa0
	$25$	$1$	$50$	$2$

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

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$	$- 1$	$- 50$	$- 2$
	$25$	$1$	$50$	$2$

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

$- 1$	$25$	$26$	$51$	$52$	$2$
	\xa0	$- 25$	$- 1$	$- 50$	$- 2$
	$25$	$1$	$50$	$2$	$0$

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

$25 x^{3} + 1 x^{2} + (50) x + 2$

Simplify the quotient polynomial.

$25 x^{3} + x^{2} + 50 x + 2$

Step 7

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (25 x^{3} + x^{2}) + 50 x + 2$

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

$(x + 1) + x^{2} (25 x + 1) + 2 (25 x + 1)$

$(x + 1) x^{2} (25 x + 1) + 2 (25 x + 1)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $25 x + 1$ .

$(x + 1) (25 x + 1) (x^{2} + 2)$

Step 9

Factor the left side of the equation.

Regroup terms.

$26 x^{3} + 52 x + 25 x^{4} + 51 x^{2} + 2 = 0$

Factor $26 x$ out of $26 x^{3} + 52 x$ .

Factor $26 x$ out of $26 x^{3}$ .

$26 x (x^{2}) + 52 x + 25 x^{4} + 51 x^{2} + 2 = 0$

Factor $26 x$ out of $52 x$ .

$26 x (x^{2}) + 26 x (2) + 25 x^{4} + 51 x^{2} + 2 = 0$

Factor $26 x$ out of $26 x (x^{2}) + 26 x (2)$ .

$26 x (x^{2} + 2) + 25 x^{4} + 51 x^{2} + 2 = 0$

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

$26 x (x^{2} + 2) + 25 {(x^{2})}^{2} + 51 x^{2} + 2 = 0$

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

$26 x (x^{2} + 2) + 25 u^{2} + 51 u + 2 = 0$

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 $a \cdot c = 25 \cdot 2 = 50$ and whose sum is $b = 51$ .

Factor $51$ out of $51 u$ .

$26 x (x^{2} + 2) + 25 u^{2} + 51 (u) + 2 = 0$

Rewrite $51$ as $1$ plus $50$

$26 x (x^{2} + 2) + 25 u^{2} + (1 + 50) u + 2 = 0$

Apply the distributive property.

$26 x (x^{2} + 2) + 25 u^{2} + 1 u + 50 u + 2 = 0$

Multiply $u$ by $1$ .

$26 x (x^{2} + 2) + 25 u^{2} + u + 50 u + 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$26 x (x^{2} + 2) + 25 u^{2} + u + 50 u + 2 = 0$

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

$26 x (x^{2} + 2) + u (25 u + 1) + 2 (25 u + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $25 u + 1$ .

$26 x (x^{2} + 2) + (25 u + 1) (u + 2) = 0$

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

$26 x (x^{2} + 2) + (25 x^{2} + 1) (x^{2} + 2) = 0$

Factor $x^{2} + 2$ out of $26 x (x^{2} + 2) + (25 x^{2} + 1) (x^{2} + 2)$ .

Factor $x^{2} + 2$ out of $26 x (x^{2} + 2)$ .

$(x^{2} + 2) (26 x) + (25 x^{2} + 1) (x^{2} + 2) = 0$

Factor $x^{2} + 2$ out of $(25 x^{2} + 1) (x^{2} + 2)$ .

$(x^{2} + 2) (26 x) + (x^{2} + 2) (25 x^{2} + 1) = 0$

Factor $x^{2} + 2$ out of $(x^{2} + 2) (26 x) + (x^{2} + 2) (25 x^{2} + 1)$ .

$(x^{2} + 2) (26 x + 25 x^{2} + 1) = 0$

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

$(x^{2} + 2) (26 u + 25 u^{2} + 1) = 0$

Factor by grouping.

Reorder terms.

$(x^{2} + 2) (25 u^{2} + 26 u + 1) = 0$

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 $a \cdot c = 25 \cdot 1 = 25$ and whose sum is $b = 26$ .

Factor $26$ out of $26 u$ .

$(x^{2} + 2) (25 u^{2} + 26 (u) + 1) = 0$

Rewrite $26$ as $1$ plus $25$

$(x^{2} + 2) (25 u^{2} + (1 + 25) u + 1) = 0$

Apply the distributive property.

$(x^{2} + 2) (25 u^{2} + 1 u + 25 u + 1) = 0$

Multiply $u$ by $1$ .

$(x^{2} + 2) (25 u^{2} + u + 25 u + 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{2} + 2) ((25 u^{2} + u) + 25 u + 1) = 0$

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

$(x^{2} + 2) (u (25 u + 1) + 1 (25 u + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $25 u + 1$ .

$(x^{2} + 2) ((25 u + 1) (u + 1)) = 0$

Factor.

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

$(x^{2} + 2) ((25 x + 1) (x + 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 2) (25 x + 1) (x + 1) = 0$

Step 10

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

$x^{2} + 2 = 0$

$25 x + 1 = 0$

$x + 1 = 0$

Step 11

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

Set $x^{2} + 2$ equal to $0$ .

$x^{2} + 2 = 0$

Solve $x^{2} + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$x^{2} = - 2$

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

$x = \pm \sqrt{- 2}$

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{2}$

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.

$x = i \sqrt{2}$

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

$x = - i \sqrt{2}$

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

$x = i \sqrt{2}, - i \sqrt{2}$

Step 12

Set $25 x + 1$ equal to $0$ and solve for $x$ .

Set $25 x + 1$ equal to $0$ .

$25 x + 1 = 0$

Solve $25 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$25 x = - 1$

Divide each term in $25 x = - 1$ by $25$ and simplify.

Divide each term in $25 x = - 1$ by $25$ .

$\frac{25 x}{25} = \frac{- 1}{25}$

Simplify the left side.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 x}{25} = \frac{- 1}{25}$

Divide $x$ by $1$ .

$x = \frac{- 1}{25}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{25}$

Step 13

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.

$x = - 1$

Step 14

The final solution is all the values that make $(x^{2} + 2) (25 x + 1) (x + 1) = 0$ true.

$x = i \sqrt{2}, - i \sqrt{2}, - \frac{1}{25}, - 1$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 25x^4+26x^3+51x^2+52x+2

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