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

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Example Question #1 : How To Solve A Polynomial In Pre Algebra

Solve $x^{3}=125$

Possible Answers:

x=15

x=25

x=75

x=5

Correct answer:

x=5

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 $\sqrt[3]{x^{3}}=\sqrt[3]{125}$

Then we perform the cube root to get the answer x=5 .

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Example Question #2 : How To Solve A Polynomial In Pre Algebra

Solve $x^{2}=64$

Possible Answers:

x=32

x=64

x=8

$x=\pm8$

Correct answer:

$x=\pm8$

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 $\sqrt{x^{}2}=\sqrt{64}$

Then we perform the square root to get the answer x=8

Remember that squared can also equal so our answer will be both positive and negative.

The final answer looks like $x=\pm8$

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Example Question #3 : How To Solve A Polynomial In Pre Algebra

Solve $x^{3}= 125$

Possible Answers:

x=25

x=15

x=12

x=5

Correct answer:

x=5

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 $\sqrt[3]{x^3}=\sqrt[3]{125}$

Then we perform the cube root to get the answer x=5 .

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Example Question #4 : How To Solve A Polynomial In Pre Algebra

Sovle $x^{2}=144$ .

Possible Answers:

x=12

$x=\pm12$

$x=\pm6$

$x=\pm4$

Correct answer:

$x=\pm12$

Explanation:

To solve an equation with $x^{2}$ we must take the square root of each side of the equation to get by itself.

Doing this makes our problem look like $\sqrt{x^{2}}=\sqrt{144}$

Then we perform the square root to get the answer $x=\pm12$

Remember that $-12^{2}$ can also equal 144 so our answer will be both positive and negative.

The final answer looks like $x=\pm12$ .

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Example Question #5 : How To Solve A Polynomial In Pre Algebra

Solve for if, $x^{4}=256$

Possible Answers:

$\pm16$

$\pm2$

$\pm4$

$\pm8$

Correct answer:

$\pm4$

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 our 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 looks like $x=\pm4$ .

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Example Question #6 : How To Solve A Polynomial In Pre Algebra

Solve for if, $x^{2}=121$

Possible Answers:

$x=\pm12$

$x=\pm13$

$x=\pm11$

$x=\pm15$

Correct answer:

$x=\pm11$

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

Then we perform the square root to get the answer x=11

Remember that $(-11)^{2}$ can also equal 121 so our answer will be both positive and negative.

The final answer looks like $x=\pm11$ .

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Example Question #7 : How To Solve A Polynomial In Pre Algebra

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

Possible Answers:

x=3

x=6

x=8

x=4

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 our problem look like $\sqrt[3]{x^{3}}=\sqrt[3]{64}$ .

Then we take the cubed root of both sides to get x=4 .

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Example Question #8 : How To Solve A Polynomial In Pre Algebra

Solve for if $x^{5}=3125$ .

Possible Answers:

x=7

x=12

x=15

x=5

Correct answer:

x=5

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 our problem look like $\sqrt[5]{x^{5}}=\sqrt[5]{3125}$ .

Then we take the fifth root to get the answer, $x=\sqrt[5]{3125}$ .

The final answer is x=5 .

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Example Question #9 : How To Solve A Polynomial In Pre Algebra

Solve for if $x^{2}=81$ .

Possible Answers:

$x=\pm6$

$x=\pm7$

$x=\pm9$

$x=\pm8$

Correct answer:

$x=\pm9$

Explanation:

To solve an equation with , we must take the square root of each side of the equation to isolate :

$\sqrt{x^{2}}=\sqrt{81}$

$x=\sqrt{81}$

Remember that $(-9)^{2}$ can also equal , so our answer needs to include both positives and negatives.

The final answer is then $x=\pm9$ .

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Example Question #1 : How To Solve A Polynomial In Pre Algebra

Solve for when $x^{4}=1296$

Possible Answers:

x=9

x=6

x=3

$x=\pm6$

Correct answer:

$x=\pm6$

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 $\sqrt[4]{x^{4}}=\sqrt[4]{1296}$

Then we perform the quad root to get the answer.

Remember that $-6^{4}$ can also equal 1296 so our answer will be both positive and negative.

The final answer is $x=\pm6$ .

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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 #1 : How To Solve A Polynomial In Pre Algebra

Example Question #2 : How To Solve A Polynomial In Pre Algebra

Example Question #3 : How To Solve A Polynomial In Pre Algebra

Example Question #4 : How To Solve A Polynomial In Pre Algebra

Example Question #5 : How To Solve A Polynomial In Pre Algebra

Example Question #6 : How To Solve A Polynomial In Pre Algebra

Example Question #7 : How To Solve A Polynomial In Pre Algebra

Example Question #8 : How To Solve A Polynomial In Pre Algebra

Example Question #9 : How To Solve A Polynomial In Pre Algebra

Example Question #1 : How To Solve A Polynomial In Pre Algebra

All High School Math Resources

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