## Algebra: A Combined Approach (4th Edition)

In a perfect square trinomial, in th form $ax^{2} + bx + c$, the constant (c) is equal to the square of half the coefficient of the second term, b. $c = (\frac{b}{2})^{2}$ Substitute c with 16 and solve for b. $16 = (\frac{b}{2})^{2}$ To undo the square, we need to take the square root of 16, which will equal to either positive 4 or negative 4. In this case, we will split this off into two equations to solve for the two possible values for the second term coefficient. $4 = \frac{b}{2}$ and $-4 = \frac{b}{2}$ Solve for b by multiplying both sides of the equation by 2. b = 8 and b = -8