Solve by Factoring 2x^2-9x+4=0

$2 x^{2} - 9 x + 4 = 0$

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 $a \cdot c = 2 \cdot 4 = 8$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$2 x^{2} - 9 x + 4 = 0$

Rewrite $- 9$ as $- 1$ plus $- 8$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Step 2

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

$2 x - 1 = 0$

$x - 4 = 0$

Step 3

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

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

Solve by Factoring 2x^2-9x+4=0

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