Solve by Factoring x^10-31x^5-32=0

$x^{10} - 31 x^{5} - 32 = 0$

Step 1

Rewrite $x^{10}$ as ${(x^{5})}^{2}$ .

${(x^{5})}^{2} - 31 x^{5} - 32 = 0$

Step 2

Let $u = x^{5}$ . Substitute $u$ for all occurrences of $x^{5}$ .

$u^{2} - 31 u - 32 = 0$

Step 3

Factor $u^{2} - 31 u - 32$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 32$ and whose sum is $- 31$ .

$- 32, 1$

Write the factored form using these integers.

$(u - 32) (u + 1) = 0$

Step 4

Replace all occurrences of $u$ with $x^{5}$ .

$(x^{5} - 32) (x^{5} + 1) = 0$

Step 5

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

$x^{5} - 32 = 0$

$x^{5} + 1 = 0$

Step 6

Set $x^{5} - 32$ equal to $0$ and solve for $x$ .

Set $x^{5} - 32$ equal to $0$ .

$x^{5} - 32 = 0$

Solve $x^{5} - 32 = 0$ for $x$ .

Add $32$ to both sides of the equation.

$x^{5} = 32$

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

$x = \sqrt[5]{32}$

Simplify $\sqrt[5]{32}$ .

Rewrite $32$ as $2^{5}$ .

$x = \sqrt[5]{2^{5}}$

Pull terms out from under the radical, assuming real numbers.

$x = 2$

Step 7

Set $x^{5} + 1$ equal to $0$ and solve for $x$ .

Set $x^{5} + 1$ equal to $0$ .

$x^{5} + 1 = 0$

Solve $x^{5} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{5} = - 1$

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

$x = \sqrt[5]{- 1}$

Simplify $\sqrt[5]{- 1}$ .

Rewrite $- 1$ as ${(- 1)}^{5}$ .

$x = \sqrt[5]{{(- 1)}^{5}}$

Pull terms out from under the radical, assuming real numbers.

$x = - 1$

Step 8

The final solution is all the values that make $(x^{5} - 32) (x^{5} + 1) = 0$ true.

$x = 2, - 1$

Solve by Factoring x^10-31x^5-32=0

Related Questions

Support LCMGCF.com by disabling your adblocker.