Find a,b,c (3x+4)(3x-4)=0

$(3 x + 4) (3 x - 4) = 0$

Step 1

Simplify the left side.

Simplify $(3 x + 4) (3 x - 4)$ .

Expand $(3 x + 4) (3 x - 4)$ using the FOIL Method.

Apply the distributive property.

$3 x (3 x - 4) + 4 (3 x - 4) = 0$

Apply the distributive property.

$3 x (3 x) + 3 x \cdot - 4 + 4 (3 x - 4) = 0$

Apply the distributive property.

$3 x (3 x) + 3 x \cdot - 4 + 4 (3 x) + 4 \cdot - 4 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 \cdot 3 x \cdot x + 3 x \cdot - 4 + 4 (3 x) + 4 \cdot - 4 = 0$

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

Move $x$ .

$3 \cdot 3 (x \cdot x) + 3 x \cdot - 4 + 4 (3 x) + 4 \cdot - 4 = 0$

Multiply $x$ by $x$ .

$3 \cdot 3 x^{2} + 3 x \cdot - 4 + 4 (3 x) + 4 \cdot - 4 = 0$

Multiply $3$ by $3$ .

$9 x^{2} + 3 x \cdot - 4 + 4 (3 x) + 4 \cdot - 4 = 0$

Multiply $- 4$ by $3$ .

$9 x^{2} - 12 x + 4 (3 x) + 4 \cdot - 4 = 0$

Multiply $3$ by $4$ .

$9 x^{2} - 12 x + 12 x + 4 \cdot - 4 = 0$

Multiply $4$ by $- 4$ .

$9 x^{2} - 12 x + 12 x - 16 = 0$

Add $- 12 x$ and $12 x$ .

$9 x^{2} + 0 - 16 = 0$

Add $9 x^{2}$ and $0$ .

$9 x^{2} - 16 = 0$

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.

$a = 9$ , $b = 0$ , $c = - 16$