How to solve a polynomial in pre-algebra

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Math › How to solve a polynomial in pre-algebra

Questions 1 - 10

1

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

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 .

2

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

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 .

3

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

$x=\pm9$

$x=\pm7$

$x=\pm8$

$x=\pm6$

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 $x=\pm9$ .

4

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

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 .

5

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

$x=\pm11$

$x=\pm12$

$x=\pm13$

$x=\pm15$

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

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

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

6

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

$\pm4$

$\pm2$

$\pm8$

$\pm16$

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

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

The final answer looks like $x=\pm4$ .

7

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

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 .

8

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

$x=\pm4$

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 .

The final answer is .

9

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

$x=\pm5$

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

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

The final answer is $x=\pm5$ .

10

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

$x=\pm 4$

$x=\pm 8$

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

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

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

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