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High School Math : How to solve a polynomial in pre-algebra

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Example Question #11 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{4}=256$ .

Possible Answers:

$x=\pm 4$

x=256

$x=\pm 8$

x=12

Correct answer:

$x=\pm 4$

Explanation:

To solve an equation with we must take the quad root of each side of the equation to get by itself.

Doing this makes the problem look like $\sqrt[4]{x^{4}}=\sqrt[4]{256}$

Then we perform the quad root to get the answer x=4

Remember that $-4^{4}$ can also equal 256 so our answer will be both positive and negative.

The final answer is $x=\pm 4$ .

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Example Question #12 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{3}=729$ .

Possible Answers:

x=7

x=11

x=13

x=9

Correct answer:

x=9

Explanation:

To solve an equation with we must take the cube root of each side of the equation to get by itself.

Doing this makes the problem look like $\sqrt[3]{x^{3}}=\sqrt[3]{729}$

Then we perform the cube root to get the answer x=9 .

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Example Question #13 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{5}=32$ ?

Possible Answers:

x=2

x=4

$x=\pm4$

x=6

Correct answer:

x=2

Explanation:

To solve an equation with we must take the fifth root of each side of the equation to get by itself.

Doing this makes the problem look like $\sqrt[5]{x^{5}}=\sqrt[5]{32}$ .

Then we perform the fifth root to get the answer x=2 .

The final answer is x=2 .

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Example Question #14 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{3}=64$ .

Possible Answers:

x=4

x=8

x=16

x=2

Correct answer:

x=4

Explanation:

To solve an equation with we must take the cube root of each side of the equation to get by itself.

Doing this makes the problem look like $\sqrt[3]{x^{3}}=\sqrt[3]{64}$

Then we perform the cube root to get the answer x=4 .

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Example Question #15 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{2}=25$ .

Possible Answers:

x=25

x=5

$x=\pm5$

x=15

Correct answer:

$x=\pm5$

Explanation:

To solve an equation with we must take the square root of each side of the equation to get by itself.

Doing this makes our problem look like $\sqrt{x^{2}}=\sqrt{25}$

Then we perform the square root to get the answer x=5

Remember that $-5^{2}$ can also equal so our answer will be both positive and negative.

The final answer is $x=\pm5$ .

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Example Question #1111 : High School Math

Solve for when $x^{3}=8$

Possible Answers:

x=4

x=2

x=8

x=3

Correct answer:

x=2

Explanation:

To solve an equation with , take the cube root of each side of the equation:

$\sqrt[3]{x^{3}}=\sqrt[3]{8}$

x=2

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Example Question #17 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{5}=3125$

Possible Answers:

x=7

x=5

x=4

x=3

Correct answer:

x=5

Explanation:

To solve an equation with , take the fifth root of both sides of the equation:

$\sqrt[5]{x^{5}}=\sqrt[5]{3125}$

x=5

We can check our answer by raising it to the fifth power:

5^5=3125

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Example Question #18 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{4}=2401$

Possible Answers:

$x=\pm7$

$x=\pm9$

$x=\pm5$

$x=\pm11$

Correct answer:

$x=\pm7$

Explanation:

To solve an equation with , take the fourth root of both sides of the equation.

$\sqrt[4]{x^{4}}=\sqrt[4]{2401}$

x=7

However, remember that $(-7)^{4}$ is also equal to 2401 , so the answer is both positive and negative .

$x=\pm7$

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Example Question #11 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{2}=121$

Possible Answers:

$x=\pm11$

$x=\pm8$

$x=\pm12$

$x=\pm9$

Correct answer:

$x=\pm11$

Explanation:

To solve an equation with , take the square root of both sides of the equatio:

$\sqrt{x^{2}}=\sqrt{121}$

x=11

However, remember that $(-11)^{2}$ is also equal to 121 , so the answer is both positive and negative .

$x=\pm11$

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Example Question #20 : How To Solve A Polynomial In Pre Algebra

Solve for if, $x^{3}=27$ .

Possible Answers:

x=3

x=13

x=6

x=9

Correct answer:

x=3

Explanation:

To solve an equation with we must take the cube root of each side of the equation to get by itself.

Doing this makes our problem look like $\sqrt[3]{x^{3}}=\sqrt[3]{27}$

Then we perform the cube root to get the answer x=3 .

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High School Math : How to solve a polynomial in pre-algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : How To Solve A Polynomial In Pre Algebra

Example Question #12 : How To Solve A Polynomial In Pre Algebra

Example Question #13 : How To Solve A Polynomial In Pre Algebra

Example Question #14 : How To Solve A Polynomial In Pre Algebra

Example Question #15 : How To Solve A Polynomial In Pre Algebra

Example Question #1111 : High School Math

Example Question #17 : How To Solve A Polynomial In Pre Algebra

Example Question #18 : How To Solve A Polynomial In Pre Algebra

Example Question #11 : How To Solve A Polynomial In Pre Algebra

Example Question #20 : How To Solve A Polynomial In Pre Algebra

All High School Math Resources

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