Solve by Factoring 4x^2+27x-7=0

$4 x^{2} + 27 x - 7 = 0$

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 $a \cdot c = 4 \cdot - 7 = - 28$ and whose sum is $b = 27$ .

Factor $27$ out of $27 x$ .

$4 x^{2} + 27 (x) - 7 = 0$

Rewrite $27$ as $- 1$ plus $28$

$4 x^{2} + (- 1 + 28) x - 7 = 0$

Apply the distributive property.

$4 x^{2} - 1 x + 28 x - 7 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{2} - 1 x) + 28 x - 7 = 0$

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

$x (4 x - 1) + 7 (4 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(4 x - 1) (x + 7) = 0$

Step 2

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

$4 x - 1 = 0$

$x + 7 = 0$

Step 3

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 4

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 5

The final solution is all the values that make $(4 x - 1) (x + 7) = 0$ true.

$x = \frac{1}{4}, - 7$

Solve by Factoring 4x^2+27x-7=0

Related Questions

Support LCMGCF.com by disabling your adblocker.