# SAT Math Multiple Choice Question 619: Answer and Explanation

Home > SAT Test > SAT Math Multiple Choice Practice Tests

### Test Information

- Use your browser's back button to return to your test results.
- Do more SAT Math Multiple Choice Practice Tests.

**Question: 619**

**4.** If *x* - 3 is a factor of the expression *x*^{2} + *kx* + 12, what is the value of *k*?

- A. -7
- B. -5
- C. 5
- D. 7

**Correct Answer:** A

**Explanation:**

**A**

**Advanced Mathematics (polynomials) EASY**

There are several ways to approach this question. Perhaps the simplest is to use the Factor Theorem: If *x* - *c* is a factor of a polynomial, then *x* = *c* is a zero of that polynomial. Therefore, if *x* - 3 is a factor of our polynomial, *x* = 3 must be a zero:

*x*^{2} + *kx* + 12 = (3)^{2} + 3*k* + 12 = 0

Simplify:

9 + 3*k* + 12 = 0

Subtract 21:

3*k* = -21

Divide by 3:

*k* = -7

Alternately, you might try to find the other factor of the quadratic. Since the constant term in the quadratic is 12, the constant term in the other binomial factor must be 12 ÷ -3 = -4.

(*x* - 3)(*x* - 4) = *x*^{2} + *kx* + 12

FOIL:

*x*^{2} - 7*x* + 12 = *x*^{2} + *kx* + 12

Subtract *x* and 12:

-7*x* = *kx*

Divide by *x*:

-7 = *k*