Solve by Factoring 5x^2-19x-4=0

$5 x^{2} - 19 x - 4 = 0$

Step 1

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 $b = - 19$ .

Factor $- 19$ out of $- 19 x$ .

$5 x^{2} - 19 x - 4 = 0$

Rewrite $- 19$ as $1$ plus $- 20$

$5 x^{2} + (1 - 20) x - 4 = 0$

Apply the distributive property.

$5 x^{2} + 1 x - 20 x - 4 = 0$

Multiply $x$ by $1$ .

$5 x^{2} + x - 20 x - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{2} + x) - 20 x - 4 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $5 x + 1$ .

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

Step 2

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

$5 x + 1 = 0$

$x - 4 = 0$

Step 3

Set $5 x + 1$ equal to $0$ and solve for $x$ .

Set $5 x + 1$ equal to $0$ .

$5 x + 1 = 0$

Solve $5 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

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.

$x = 4$

Step 5

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

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

Solve by Factoring 5x^2-19x-4=0

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