Solve for x 6x-x^2-8=0

$6 x - x^{2} - 8 = 0$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$6 u - u^{2} - 8 = 0$

Factor by grouping.

Reorder terms.

$- u^{2} + 6 u - 8 = 0$

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 $a \cdot c = - 1 \cdot - 8 = 8$ and whose sum is $b = 6$ .

Factor $6$ out of $6 u$ .

$- u^{2} + 6 (u) - 8 = 0$

Rewrite $6$ as $2$ plus $4$

$- u^{2} + (2 + 4) u - 8 = 0$

Apply the distributive property.

$- u^{2} + 2 u + 4 u - 8 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(- u^{2} + 2 u) + 4 u - 8 = 0$

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

$u (- u + 2) - 4 (- u + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $- u + 2$ .

$(- u + 2) (u - 4) = 0$

Replace all occurrences of $u$ with $x$ .

$(- x + 2) (x - 4) = 0$

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

$- x + 2 = 0$

$x - 4 = 0$

Set $- x + 2$ equal to $0$ and solve for $x$ .

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Solve $- x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(- x + 2) (x - 4) = 0$ true.

$x = 2, 4$

Solve for x 6x-x^2-8=0

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