Basic Math Examples

$5 - 4 q = 1 q^{2}$

Step 1

Multiply $q^{2}$ by $1$ .

$5 - 4 q = q^{2}$

Step 2

Subtract $q^{2}$ from both sides of the equation.

$5 - 4 q - q^{2} = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $- 1$ out of $5 - 4 q - q^{2}$ .

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Step 3.1.1

Reorder the expression.

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Step 3.1.1.1

Move $5$ .

$- 4 q - q^{2} + 5 = 0$

Step 3.1.1.2

Reorder $- 4 q$ and $- q^{2}$ .

$- q^{2} - 4 q + 5 = 0$

Step 3.1.2

Factor $- 1$ out of $- q^{2}$ .

$- (q^{2}) - 4 q + 5 = 0$

Step 3.1.3

Factor $- 1$ out of $- 4 q$ .

$- (q^{2}) - (4 q) + 5 = 0$

Step 3.1.4

Rewrite $5$ as $- 1 (- 5)$ .

$- (q^{2}) - (4 q) - 1 \cdot - 5 = 0$

Step 3.1.5

Factor $- 1$ out of $- (q^{2}) - (4 q)$ .

$- (q^{2} + 4 q) - 1 \cdot - 5 = 0$

Step 3.1.6

Factor $- 1$ out of $- (q^{2} + 4 q) - 1 (- 5)$ .

$- (q^{2} + 4 q - 5) = 0$

Step 3.2

Factor.

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Step 3.2.1

Factor $q^{2} + 4 q - 5$ using the AC method.

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Step 3.2.1.1

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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 3.2.1.2

Write the factored form using these integers.

$- ((q - 1) (q + 5)) = 0$

Step 3.2.2

Remove unnecessary parentheses.

$- (q - 1) (q + 5) = 0$

Step 4

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

$q - 1 = 0$

$q + 5 = 0$

Step 5

Set $q - 1$ equal to $0$ and solve for $q$ .

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Step 5.1

Set $q - 1$ equal to $0$ .

$q - 1 = 0$

Step 5.2

Add $1$ to both sides of the equation.

$q = 1$

Step 6

Set $q + 5$ equal to $0$ and solve for $q$ .

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Step 6.1

Set $q + 5$ equal to $0$ .

$q + 5 = 0$

Step 6.2

Subtract $5$ from both sides of the equation.

$q = - 5$

Step 7

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

$q = 1, - 5$