Solve by Factoring 21-21x=(5x+4)(x-1)

$21 - 21 x = (5 x + 4) (x - 1)$

Step 1

Subtract $(5 x + 4) (x - 1)$ from both sides of the equation.

$21 - 21 x - (5 x + 4) (x - 1) = 0$

Step 2

Simplify $21 - 21 x - (5 x + 4) (x - 1)$ .

Simplify each term.

Apply the distributive property.

$21 - 21 x + (- (5 x) - 1 \cdot 4) (x - 1) = 0$

Multiply $5$ by $- 1$ .

$21 - 21 x + (- 5 x - 1 \cdot 4) (x - 1) = 0$

Multiply $- 1$ by $4$ .

$21 - 21 x + (- 5 x - 4) (x - 1) = 0$

Expand $(- 5 x - 4) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$21 - 21 x - 5 x (x - 1) - 4 (x - 1) = 0$

Apply the distributive property.

$21 - 21 x - 5 x \cdot x - 5 x \cdot - 1 - 4 (x - 1) = 0$

Apply the distributive property.

$21 - 21 x - 5 x \cdot x - 5 x \cdot - 1 - 4 x - 4 \cdot - 1 = 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$21 - 21 x - 5 (x \cdot x) - 5 x \cdot - 1 - 4 x - 4 \cdot - 1 = 0$

Multiply $x$ by $x$ .

$21 - 21 x - 5 x^{2} - 5 x \cdot - 1 - 4 x - 4 \cdot - 1 = 0$

Multiply $- 1$ by $- 5$ .

$21 - 21 x - 5 x^{2} + 5 x - 4 x - 4 \cdot - 1 = 0$

Multiply $- 4$ by $- 1$ .

$21 - 21 x - 5 x^{2} + 5 x - 4 x + 4 = 0$

Subtract $4 x$ from $5 x$ .

$21 - 21 x - 5 x^{2} + x + 4 = 0$

Add $21$ and $4$ .

$- 21 x - 5 x^{2} + x + 25 = 0$

Add $- 21 x$ and $x$ .

$- 5 x^{2} - 20 x + 25 = 0$

Step 3

Factor $5$ out of $- 5 x^{2} - 20 x + 25$ .

Factor $5$ out of $- 5 x^{2}$ .

$5 (- x^{2}) - 20 x + 25 = 0$

Factor $5$ out of $- 20 x$ .

$5 (- x^{2}) + 5 (- 4 x) + 25 = 0$

Factor $5$ out of $25$ .

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

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

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

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

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

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 $a \cdot c = - 1 \cdot 5 = - 5$ and whose sum is $b = - 4$ .

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

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

Rewrite $- 4$ as $1$ plus $- 5$

$5 (- x^{2} + (1 - 5) x + 5) = 0$

Apply the distributive property.

$5 (- x^{2} + 1 x - 5 x + 5) = 0$

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 5

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

$- x + 1 = 0$

$x + 5 = 0$

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 $x + 5$ equal to $0$ and solve for $x$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

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

$x = 1, - 5$

Solve by Factoring 21-21x=(5x+4)(x-1)

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