Pre-Algebra Examples

$26 x (x + 3) = 8840$

Step 1

Divide each term in $26 x (x + 3) = 8840$ by $26$ and simplify.

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Step 1.1

Divide each term in $26 x (x + 3) = 8840$ by $26$ .

$\frac{26 x (x + 3)}{26} = \frac{8840}{26}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $26$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{26 x (x + 3)}{26} = \frac{8840}{26}$

Step 1.2.1.2

Divide $x (x + 3)$ by $1$ .

$x (x + 3) = \frac{8840}{26}$

Step 1.2.2

Apply the distributive property.

$x \cdot x + x \cdot 3 = \frac{8840}{26}$

Step 1.2.3

Simplify the expression.

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Step 1.2.3.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot 3 = \frac{8840}{26}$

Step 1.2.3.2

Move $3$ to the left of $x$ .

$x^{2} + 3 x = \frac{8840}{26}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $8840$ by $26$ .

$x^{2} + 3 x = 340$

Step 2

Subtract $340$ from both sides of the equation.

$x^{2} + 3 x - 340 = 0$

Step 3

Factor $x^{2} + 3 x - 340$ 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 $- 340$ and whose sum is $3$ .

$- 17, 20$

Step 3.2

Write the factored form using these integers.

$(x - 17) (x + 20) = 0$

Step 4

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

$x - 17 = 0$

$x + 20 = 0$

Step 5

Set $x - 17$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 17$ equal to $0$ .

$x - 17 = 0$

Step 5.2

Add $17$ to both sides of the equation.

$x = 17$

Step 6

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

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Step 6.1

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

$x + 20 = 0$

Step 6.2

Subtract $20$ from both sides of the equation.

$x = - 20$

Step 7

The final solution is all the values that make $(x - 17) (x + 20) = 0$ true.

$x = 17, - 20$