Find a,b,c (2x+7)(x-1)=0

$(2 x + 7) (x - 1) = 0$

Step 1

Simplify the left side.

Simplify $(2 x + 7) (x - 1)$ .

Expand $(2 x + 7) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$2 x (x - 1) + 7 (x - 1) = 0$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 1 + 7 (x - 1) = 0$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 1 + 7 x + 7 \cdot - 1 = 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 1 + 7 x + 7 \cdot - 1 = 0$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 1 + 7 x + 7 \cdot - 1 = 0$

Multiply $- 1$ by $2$ .

$2 x^{2} - 2 x + 7 x + 7 \cdot - 1 = 0$

Multiply $7$ by $- 1$ .

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

Add $- 2 x$ and $7 x$ .

$2 x^{2} + 5 x - 7 = 0$

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.

$a = 2$ , $b = 5$ , $c = - 7$