Pre-Algebra Examples

$(x + 4) (x - 15) = 6 x$

Step 1

Simplify $(x + 4) (x - 15)$ .

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Step 1.1

Rewrite.

$0 + 0 + (x + 4) (x - 15) = 6 x$

Step 1.2

Simplify by adding zeros.

$(x + 4) (x - 15) = 6 x$

Step 1.3

Expand $(x + 4) (x - 15)$ using the FOIL Method.

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Step 1.3.1

Apply the distributive property.

$x (x - 15) + 4 (x - 15) = 6 x$

Step 1.3.2

Apply the distributive property.

$x \cdot x + x \cdot - 15 + 4 (x - 15) = 6 x$

Step 1.3.3

Apply the distributive property.

$x \cdot x + x \cdot - 15 + 4 x + 4 \cdot - 15 = 6 x$

Step 1.4

Simplify and combine like terms.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 15 + 4 x + 4 \cdot - 15 = 6 x$

Step 1.4.1.2

Move $- 15$ to the left of $x$ .

$x^{2} - 15 \cdot x + 4 x + 4 \cdot - 15 = 6 x$

Step 1.4.1.3

Multiply $4$ by $- 15$ .

$x^{2} - 15 x + 4 x - 60 = 6 x$

Step 1.4.2

Add $- 15 x$ and $4 x$ .

$x^{2} - 11 x - 60 = 6 x$

Step 2

Move all terms containing $x$ to the left side of the equation.

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Step 2.1

Subtract $6 x$ from both sides of the equation.

$x^{2} - 11 x - 60 - 6 x = 0$

Step 2.2

Subtract $6 x$ from $- 11 x$ .

$x^{2} - 17 x - 60 = 0$

Step 3

Factor $x^{2} - 17 x - 60$ using the AC method.

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Step 3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 60$ and whose sum is $- 17$ .

$- 20, 3$

Step 3.2

Write the factored form using these integers.

$(x - 20) (x + 3) = 0$

Step 4

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

$x - 20 = 0$

$x + 3 = 0$

Step 5

Set $x - 20$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 20$ equal to $0$ .

$x - 20 = 0$

Step 5.2

Add $20$ to both sides of the equation.

$x = 20$

Step 6

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

The final solution is all the values that make $(x - 20) (x + 3) = 0$ true.

$x = 20, - 3$