Solve by Factoring 5x^2+x+1=5

$5 x^{2} + x + 1 = 5$

Subtract $5$ from both sides of the equation.

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

Subtract $5$ from $1$ .

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

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 $a \cdot c = 5 \cdot - 4 = - 20$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $5 x - 4$ .

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

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

$5 x - 4 = 0$

$x + 1 = 0$

Set $5 x - 4$ equal to $0$ and solve for $x$ .

Set $5 x - 4$ equal to $0$ .

$5 x - 4 = 0$

Solve $5 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$5 x = 4$

Divide each term in $5 x = 4$ by $5$ and simplify.

Divide each term in $5 x = 4$ by $5$ .

$\frac{5 x}{5} = \frac{4}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{4}{5}$

Divide $x$ by $1$ .

$x = \frac{4}{5}$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(5 x - 4) (x + 1) = 0$ true.

$x = \frac{4}{5}, - 1$

Solve by Factoring 5x^2+x+1=5

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