Finite Math Examples

$27 = 5 (v + 2) (v - 2) - 26 (v - 2)$

Step 1

Rewrite the equation as $5 (v + 2) (v - 2) - 26 (v - 2) = 27$ .

$5 (v + 2) (v - 2) - 26 (v - 2) = 27$

Step 2

Simplify $5 (v + 2) (v - 2) - 26 (v - 2)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$(5 v + 5 \cdot 2) (v - 2) - 26 (v - 2) = 27$

Step 2.1.2

Multiply $5$ by $2$ .

$(5 v + 10) (v - 2) - 26 (v - 2) = 27$

Step 2.1.3

Expand $(5 v + 10) (v - 2)$ using the FOIL Method.

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Step 2.1.3.1

Apply the distributive property.

$5 v (v - 2) + 10 (v - 2) - 26 (v - 2) = 27$

Step 2.1.3.2

Apply the distributive property.

$5 v \cdot v + 5 v \cdot - 2 + 10 (v - 2) - 26 (v - 2) = 27$

Step 2.1.3.3

Apply the distributive property.

$5 v \cdot v + 5 v \cdot - 2 + 10 v + 10 \cdot - 2 - 26 (v - 2) = 27$

Step 2.1.4

Simplify and combine like terms.

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Step 2.1.4.1

Simplify each term.

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Step 2.1.4.1.1

Multiply $v$ by $v$ by adding the exponents.

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Step 2.1.4.1.1.1

Move $v$ .

$5 (v \cdot v) + 5 v \cdot - 2 + 10 v + 10 \cdot - 2 - 26 (v - 2) = 27$

Step 2.1.4.1.1.2

Multiply $v$ by $v$ .

$5 v^{2} + 5 v \cdot - 2 + 10 v + 10 \cdot - 2 - 26 (v - 2) = 27$

Step 2.1.4.1.2

Multiply $- 2$ by $5$ .

$5 v^{2} - 10 v + 10 v + 10 \cdot - 2 - 26 (v - 2) = 27$

Step 2.1.4.1.3

Multiply $10$ by $- 2$ .

$5 v^{2} - 10 v + 10 v - 20 - 26 (v - 2) = 27$

Step 2.1.4.2

Add $- 10 v$ and $10 v$ .

$5 v^{2} + 0 - 20 - 26 (v - 2) = 27$

Step 2.1.4.3

Add $5 v^{2}$ and $0$ .

$5 v^{2} - 20 - 26 (v - 2) = 27$

Step 2.1.5

Apply the distributive property.

$5 v^{2} - 20 - 26 v - 26 \cdot - 2 = 27$

Step 2.1.6

Multiply $- 26$ by $- 2$ .

$5 v^{2} - 20 - 26 v + 52 = 27$

Step 2.2

Add $- 20$ and $52$ .

$5 v^{2} - 26 v + 32 = 27$

Step 3

Subtract $27$ from both sides of the equation.

$5 v^{2} - 26 v + 32 - 27 = 0$

Step 4

Subtract $27$ from $32$ .

$5 v^{2} - 26 v + 5 = 0$

Step 5

Factor by grouping.

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Step 5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot 5 = 25$ and whose sum is $b = - 26$ .

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Step 5.1.1

Factor $- 26$ out of $- 26 v$ .

$5 v^{2} - 26 v + 5 = 0$

Step 5.1.2

Rewrite $- 26$ as $- 1$ plus $- 25$

$5 v^{2} + (- 1 - 25) v + 5 = 0$

Step 5.1.3

Apply the distributive property.

$5 v^{2} - 1 v - 25 v + 5 = 0$

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

$(5 v^{2} - 1 v) - 25 v + 5 = 0$

Step 5.2.2

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

$v (5 v - 1) - 5 (5 v - 1) = 0$

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $5 v - 1$ .

$(5 v - 1) (v - 5) = 0$

Step 6

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

$5 v - 1 = 0$

$v - 5 = 0$

Step 7

Set $5 v - 1$ equal to $0$ and solve for $v$ .

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Step 7.1

Set $5 v - 1$ equal to $0$ .

$5 v - 1 = 0$

Step 7.2

Solve $5 v - 1 = 0$ for $v$ .

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Step 7.2.1

Add $1$ to both sides of the equation.

$5 v = 1$

Step 7.2.2

Divide each term in $5 v = 1$ by $5$ and simplify.

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Step 7.2.2.1

Divide each term in $5 v = 1$ by $5$ .

$\frac{5 v}{5} = \frac{1}{5}$

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Cancel the common factor of $5$ .

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Step 7.2.2.2.1.1

Cancel the common factor.

$\frac{5 v}{5} = \frac{1}{5}$

Step 7.2.2.2.1.2

Divide $v$ by $1$ .

$v = \frac{1}{5}$

Step 8

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

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Step 8.1

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

$v - 5 = 0$

Step 8.2

Add $5$ to both sides of the equation.

$v = 5$

Step 9

The final solution is all the values that make $(5 v - 1) (v - 5) = 0$ true.

$v = \frac{1}{5}, 5$

Step 10

The result can be shown in multiple forms.

Exact Form:

$v = \frac{1}{5}, 5$

Decimal Form:

$v = 0.2, 5$