Solve by Factoring 8x^2-6x+1=0

$8 x^{2} - 6 x + 1 = 0$

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

Factor $- 6$ out of $- 6 x$ .

$8 x^{2} - 6 x + 1 = 0$

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

$8 x^{2} + (- 2 - 4) x + 1 = 0$

Apply the distributive property.

$8 x^{2} - 2 x - 4 x + 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(8 x^{2} - 2 x) - 4 x + 1 = 0$

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

$2 x (4 x - 1) - (4 x - 1) = 0$

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

$(4 x - 1) (2 x - 1) = 0$

Step 2

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

$4 x - 1 = 0$

$2 x - 1 = 0$

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 $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 5

The final solution is all the values that make $(4 x - 1) (2 x - 1) = 0$ true.

$x = \frac{1}{4}, \frac{1}{2}$

Solve by Factoring 8x^2-6x+1=0

Related Questions

Support LCMGCF.com by disabling your adblocker.