Pre-Algebra Examples

$175 - 10 c - 5 c^{2}$

Step 1

Factor out the GCF of $5$ from $175 - 10 c - 5 c^{2}$ .

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Step 1.1

Factor out the GCF of $5$ from each term in the polynomial.

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Step 1.1.1

Factor out the GCF of $5$ from the expression $175$ .

$5 (35) - 10 c - 5 c^{2}$

Step 1.1.2

Factor out the GCF of $5$ from the expression $- 10 c$ .

$5 (35) + 5 (- 2 c) - 5 c^{2}$

Step 1.1.3

Factor out the GCF of $5$ from the expression $- 5 c^{2}$ .

$5 (35) + 5 (- 2 c) + 5 (- c^{2})$

Step 1.2

Since all the terms share a common factor of $5$ , it can be factored out of each term.

$5 (35 - 2 c - c^{2})$

Step 2

Rewrite $35$ as $36$ plus $- 1$

$5 (36 - 1 - 2 c - c^{2})$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $1$ as $1^{2}$ .

$5 (36 - (1^{2} + 2 c + c^{2}))$

Step 3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 c = 2 \cdot 1 \cdot c$

Step 3.3

Rewrite the polynomial.

$5 (36 - (1^{2} + 2 \cdot 1 \cdot c + c^{2}))$

Step 3.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 1$ and $b = c$ .

$5 (36 - {(1 + c)}^{2})$

Step 4

Rewrite $36$ as $6^{2}$ .

$5 (6^{2} - {(1 + c)}^{2})$

Step 5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 6$ and $b = 1 + c$ .

$5 ((6 + 1 + c) (6 - (1 + c)))$

Step 6

Simplify.

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Step 6.1

Add $6$ and $1$ .

$5 ((7 + c) (6 - (1 + c)))$

Step 6.2

Apply the distributive property.

$5 ((7 + c) (6 - 1 \cdot 1 - c))$

Step 6.3

Multiply $- 1$ by $1$ .

$5 ((7 + c) (6 - 1 - c))$

Step 6.4

Subtract $1$ from $6$ .

$5 ((7 + c) (5 - c))$