Basic Math Examples

$c^{3} - 4 c^{2} - 16 c + 64 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor out the greatest common factor from each group.

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Step 1.1.1

Group the first two terms and the last two terms.

$(c^{3} - 4 c^{2}) - 16 c + 64 = 0$

Step 1.1.2

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

$c^{2} (c - 4) - 16 (c - 4) = 0$

Step 1.2

Factor the polynomial by factoring out the greatest common factor, $c - 4$ .

$(c - 4) (c^{2} - 16) = 0$

Step 1.3

Rewrite $16$ as $4^{2}$ .

$(c - 4) (c^{2} - 4^{2}) = 0$

Step 1.4

Factor.

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Step 1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = c$ and $b = 4$ .

$(c - 4) ((c + 4) (c - 4)) = 0$

Step 1.4.2

Remove unnecessary parentheses.

$(c - 4) (c + 4) (c - 4) = 0$

Step 1.5

Combine exponents.

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Step 1.5.1

Raise $c - 4$ to the power of $1$ .

$(c - 4) (c - 4) (c + 4) = 0$

Step 1.5.2

Raise $c - 4$ to the power of $1$ .

$(c - 4) (c - 4) (c + 4) = 0$

Step 1.5.3

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

${(c - 4)}^{1 + 1} (c + 4) = 0$

Step 1.5.4

Add $1$ and $1$ .

${(c - 4)}^{2} (c + 4) = 0$

Step 2

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

${(c - 4)}^{2} = 0$

$c + 4 = 0$

Step 3

Set ${(c - 4)}^{2}$ equal to $0$ and solve for $c$ .

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Step 3.1

Set ${(c - 4)}^{2}$ equal to $0$ .

${(c - 4)}^{2} = 0$

Step 3.2

Solve ${(c - 4)}^{2} = 0$ for $c$ .

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Step 3.2.1

Set the $c - 4$ equal to $0$ .

$c - 4 = 0$

Step 3.2.2

Add $4$ to both sides of the equation.

$c = 4$

Step 4

Set $c + 4$ equal to $0$ and solve for $c$ .

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Step 4.1

Set $c + 4$ equal to $0$ .

$c + 4 = 0$

Step 4.2

Subtract $4$ from both sides of the equation.

$c = - 4$

Step 5

The final solution is all the values that make ${(c - 4)}^{2} (c + 4) = 0$ true.

$c = 4, - 4$