Solve by Factoring 15-15x=(4x+1)(x-1)

$15 - 15 x = (4 x + 1) (x - 1)$

Step 1

Subtract $(4 x + 1) (x - 1)$ from both sides of the equation.

$15 - 15 x - (4 x + 1) (x - 1) = 0$

Step 2

Simplify $15 - 15 x - (4 x + 1) (x - 1)$ .

Simplify each term.

Apply the distributive property.

$15 - 15 x + (- (4 x) - 1 \cdot 1) (x - 1) = 0$

Multiply $4$ by $- 1$ .

$15 - 15 x + (- 4 x - 1 \cdot 1) (x - 1) = 0$

Multiply $- 1$ by $1$ .

$15 - 15 x + (- 4 x - 1) (x - 1) = 0$

Expand $(- 4 x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$15 - 15 x - 4 x (x - 1) - 1 (x - 1) = 0$

Apply the distributive property.

$15 - 15 x - 4 x \cdot x - 4 x \cdot - 1 - 1 (x - 1) = 0$

Apply the distributive property.

$15 - 15 x - 4 x \cdot x - 4 x \cdot - 1 - 1 x - 1 \cdot - 1 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$15 - 15 x - 4 (x \cdot x) - 4 x \cdot - 1 - 1 x - 1 \cdot - 1 = 0$

Multiply $x$ by $x$ .

$15 - 15 x - 4 x^{2} - 4 x \cdot - 1 - 1 x - 1 \cdot - 1 = 0$

Multiply $- 1$ by $- 4$ .

$15 - 15 x - 4 x^{2} + 4 x - 1 x - 1 \cdot - 1 = 0$

Rewrite $- 1 x$ as $- x$ .

$15 - 15 x - 4 x^{2} + 4 x - x - 1 \cdot - 1 = 0$

Multiply $- 1$ by $- 1$ .

$15 - 15 x - 4 x^{2} + 4 x - x + 1 = 0$

Subtract $x$ from $4 x$ .

$15 - 15 x - 4 x^{2} + 3 x + 1 = 0$

Add $15$ and $1$ .

$- 15 x - 4 x^{2} + 3 x + 16 = 0$

Add $- 15 x$ and $3 x$ .

$- 4 x^{2} - 12 x + 16 = 0$

Step 3

Factor $4$ out of $- 4 x^{2} - 12 x + 16$ .

Factor $4$ out of $- 4 x^{2}$ .

$4 (- x^{2}) - 12 x + 16 = 0$

Factor $4$ out of $- 12 x$ .

$4 (- x^{2}) + 4 (- 3 x) + 16 = 0$

Factor $4$ out of $16$ .

$4 (- x^{2}) + 4 (- 3 x) + 4 (4) = 0$

Factor $4$ out of $4 (- x^{2}) + 4 (- 3 x)$ .

$4 (- x^{2} - 3 x) + 4 (4) = 0$

Factor $4$ out of $4 (- x^{2} - 3 x) + 4 (4)$ .

$4 (- x^{2} - 3 x + 4) = 0$

Step 4

Factor.

Factor by grouping.

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 $a \cdot c = - 1 \cdot 4 = - 4$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$4 (- x^{2} - 3 x + 4) = 0$

Rewrite $- 3$ as $1$ plus $- 4$

$4 (- x^{2} + (1 - 4) x + 4) = 0$

Apply the distributive property.

$4 (- x^{2} + 1 x - 4 x + 4) = 0$

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$4 (x (- x + 1) + 4 (- x + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$4 ((- x + 1) (x + 4)) = 0$

Remove unnecessary parentheses.

$4 (- x + 1) (x + 4) = 0$

Step 5

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

$- x + 1 = 0$

$x + 4 = 0$

Step 6

Set $- x + 1$ equal to $0$ and solve for $x$ .

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Solve $- x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Step 7

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8

The final solution is all the values that make $4 (- x + 1) (x + 4) = 0$ true.

$x = 1, - 4$

Solve by Factoring 15-15x=(4x+1)(x-1)

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