Find a,b,c (2x+1)(x-2)=0

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

Step 1

Simplify the left side.

Simplify $(2 x + 1) (x - 2)$ .

Expand $(2 x + 1) (x - 2)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 2 + 1 (x - 2) = 0$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 2 + 1 x + 1 \cdot - 2 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 2 + 1 x + 1 \cdot - 2 = 0$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 2 + 1 x + 1 \cdot - 2 = 0$

Multiply $- 2$ by $2$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 2$ by $1$ .

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

Add $- 4 x$ and $x$ .

$2 x^{2} - 3 x - 2 = 0$

Step 2

Once the quadratic is in standard form, the values of $a$ , $b$ , and $c$ can be found.

$a x^{2} + b x + c$

Step 3

Use the standard form of the equation to find $a$ , $b$ , and $c$ for this quadratic.

$a = 2$ , $b = - 3$ , $c = - 2$