Solve for y y^3+5y^2-25y-125=0

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

Step 1

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $y + 5$ .

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

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

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

Factor.

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

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

Remove unnecessary parentheses.

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

Combine exponents.

Raise $y + 5$ to the power of $1$ .

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

Raise $y + 5$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Step 2

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

${(y + 5)}^{2} = 0$

$y - 5 = 0$

Step 3

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

Set ${(y + 5)}^{2}$ equal to $0$ .

${(y + 5)}^{2} = 0$

Solve ${(y + 5)}^{2} = 0$ for $y$ .

Set the $y + 5$ equal to $0$ .

$y + 5 = 0$

Subtract $5$ from both sides of the equation.

$y = - 5$

Step 4

Set $y - 5$ equal to $0$ and solve for $y$ .

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

Add $5$ to both sides of the equation.

$y = 5$

Step 5

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

$y = - 5, 5$

Solve for y y^3+5y^2-25y-125=0

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