Algebra Examples

$\frac{2}{3} \cdot ((x - 4) (x + 5)) = 1$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Simplify the left side.

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Step 1.1.1

Simplify $\frac{2}{3} \cdot ((x - 4) (x + 5))$ .

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Step 1.1.1.1

Expand $(x - 4) (x + 5)$ using the FOIL Method.

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Step 1.1.1.1.1

Apply the distributive property.

$\frac{2}{3} \cdot (x (x + 5) - 4 (x + 5)) = 1$

Step 1.1.1.1.2

Apply the distributive property.

$\frac{2}{3} \cdot (x \cdot x + x \cdot 5 - 4 (x + 5)) = 1$

Step 1.1.1.1.3

Apply the distributive property.

$\frac{2}{3} \cdot (x \cdot x + x \cdot 5 - 4 x - 4 \cdot 5) = 1$

Step 1.1.1.2

Simplify and combine like terms.

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Step 1.1.1.2.1

Simplify each term.

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Step 1.1.1.2.1.1

Multiply $x$ by $x$ .

$\frac{2}{3} \cdot (x^{2} + x \cdot 5 - 4 x - 4 \cdot 5) = 1$

Step 1.1.1.2.1.2

Move $5$ to the left of $x$ .

$\frac{2}{3} \cdot (x^{2} + 5 \cdot x - 4 x - 4 \cdot 5) = 1$

Step 1.1.1.2.1.3

Multiply $- 4$ by $5$ .

$\frac{2}{3} \cdot (x^{2} + 5 x - 4 x - 20) = 1$

Step 1.1.1.2.2

Subtract $4 x$ from $5 x$ .

$\frac{2}{3} \cdot (x^{2} + x - 20) = 1$

Step 1.1.1.3

Apply the distributive property.

$\frac{2}{3} x^{2} + \frac{2}{3} x + \frac{2}{3} \cdot - 20 = 1$

Step 1.1.1.4

Simplify.

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Step 1.1.1.4.1

Combine $\frac{2}{3}$ and $x^{2}$ .

$\frac{2 x^{2}}{3} + \frac{2}{3} x + \frac{2}{3} \cdot - 20 = 1$

Step 1.1.1.4.2

Combine $\frac{2}{3}$ and $x$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} + \frac{2}{3} \cdot - 20 = 1$

Step 1.1.1.4.3

Multiply $\frac{2}{3} \cdot - 20$ .

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Step 1.1.1.4.3.1

Combine $\frac{2}{3}$ and $- 20$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} + \frac{2 \cdot - 20}{3} = 1$

Step 1.1.1.4.3.2

Multiply $2$ by $- 20$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} + \frac{- 40}{3} = 1$

Step 1.1.1.5

Move the negative in front of the fraction.

$\frac{2 x^{2}}{3} + \frac{2 x}{3} - \frac{40}{3} = 1$

Step 1.2

Subtract $1$ from both sides of the equation.

$\frac{2 x^{2}}{3} + \frac{2 x}{3} - \frac{40}{3} - 1 = 0$

Step 1.3

Simplify $\frac{2 x^{2}}{3} + \frac{2 x}{3} - \frac{40}{3} - 1$ .

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Step 1.3.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} - \frac{40}{3} - 1 \cdot \frac{3}{3} = 0$

Step 1.3.2

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} - \frac{40}{3} + \frac{- 1 \cdot 3}{3} = 0$

Step 1.3.3

Combine the numerators over the common denominator.

$\frac{2 x^{2}}{3} + \frac{2 x}{3} + \frac{- 40 - 1 \cdot 3}{3} = 0$

Step 1.3.4

Simplify the numerator.

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Step 1.3.4.1

Multiply $- 1$ by $3$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} + \frac{- 40 - 3}{3} = 0$

Step 1.3.4.2

Subtract $3$ from $- 40$ .

$\frac{2 x^{2}}{3} + \frac{2 x}{3} + \frac{- 43}{3} = 0$

Step 1.3.5

Move the negative in front of the fraction.

$\frac{2 x^{2}}{3} + \frac{2 x}{3} - \frac{43}{3} = 0$

Step 2

Once the quadratic is in standard form, the values of $a$ , $b$ , and $c$ can be found.

$a x^{2} + b x + c$

Step 3

Use the standard form of the equation to find $a$ , $b$ , and $c$ for this quadratic.

$a = \frac{2}{3}$ , $b = \frac{2}{3}$ , $c = - \frac{43}{3}$