# Practice GMAT Problem Solving Question

f(x) = x2 + 4x + k = 12; f(-6) = 0; if k is a constant and n is the number for which f(n) = 0, what is the value of n?
 A) 6 B) 2 C) 0 D) -2 E) -12
1. There are two main approaches you can take to solving this problem. You can either solve for k and turn this problem into a simple factoring problem or you can ignore k and reverse factor.
2. Method 1: Solve for K.
1. f(x) = x2 + 4x + k = 12
f(x) = x2 + 4x + k - 12 = 0 (subtract 12)
2. Plug in x=-6 knowing that the value of f(x) must equal zero since f(-6)=0.
f(-6) = (-6)2 + 4(-6) + k - 12 = 0
36 -24 + k - 12 = 0
0 - k = 0 --> k = 0
3. The equation is now:
f(x) = x2 + 4x + 0 - 12 = 0
f(x) = x2 + 4x - 12 = 0
4. By factoring: x2 + 4x - 12 = 0 equals:
(x+6)(x+n) Note: x+6 is from f(-6)=0
5. Since we need two numbers that add to +4 and multiply to -12, we know that +6 and -2 work. Consequently, (x-2) is a factor and therefore, f(+2)=0, so n = 2.
3. Method 2: Ignore K.
1. A crucial insight in unlocking this problem is recognizing that if f(a) = 0 and f(x) is a quadratic equation in the form x2 + bx + c = 0, (x – a) is a factor of the equation. In order to get the equation into quadratic form, subtract 12.
f(x) = x2 + 4x + k = 12
f(x) = x2 + 4x + k – 12 = 0
2. Since f(-6) = 0, (x+6) is a factor.
3. It is important to remember how factoring works. Specifically, remember the following:
(x + d)(x + e) = x2 + dex + de = 0
So: (x + 6)(x + a) = x2 + 4x + (k – 12)
4. With this in mind, you know that 6 + a = 4. So, a = -2 and therefore, (x - 2) is the other factor of the quadratic. So, f(2) = 0 and n = 2.