Solve by Factoring x^3-5x^2-25x+125=0

$x^{3} - 5 x^{2} - 25 x + 125 = 0$

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 5 x^{2}) - 25 x + 125 = 0$

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

$x^{2} (x - 5) - 25 (x - 5) = 0$

Step 2

Factor the polynomial by factoring out the greatest common factor, $x - 5$ .

$(x - 5) (x^{2} - 25) = 0$

Step 3

Rewrite $25$ as $5^{2}$ .

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

Step 4

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$(x - 5) ((x + 5) (x - 5)) = 0$

Remove unnecessary parentheses.

$(x - 5) (x + 5) (x - 5) = 0$

Step 5

Combine exponents.

Raise $x - 5$ to the power of $1$ .

$(x - 5) (x - 5) (x + 5) = 0$

Raise $x - 5$ to the power of $1$ .

$(x - 5) (x - 5) (x + 5) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Step 6

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

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

$x + 5 = 0$

Step 7

Set ${(x - 5)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 5)}^{2}$ equal to $0$ .

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

Solve ${(x - 5)}^{2} = 0$ for $x$ .

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 8

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 9

The final solution is all the values that make ${(x - 5)}^{2} (x + 5) = 0$ true.

$x = 5, - 5$

Solve by Factoring x^3-5x^2-25x+125=0

Related Questions

Support LCMGCF.com by disabling your adblocker.