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Math · Advanced Math

Polynomials and zeros — Digital SAT Math practice

Factors, remainders, zeros and polynomial identities.

Key terms

Common traps

Key ideas to know

Factor Theorem Application

If x − 3 is a factor of p(x), what must be true about p(3)?

p(3) must equal 0. The Factor Theorem states that x − r is a factor exactly when p(r) = 0.

Watch out: Assuming p(r) = 1 or any nonzero value means x − r is still a factor.

Using the Remainder Theorem

How can you quickly find the remainder when p(x) = 2x3 − 4x + 1 is divided by x + 2 without long division?

Evaluate p(−2). The Remainder Theorem says dividing by x − r leaves a remainder of p(r). Here r = −2, so calculate p(−2).

Watch out: Forgetting to flip the sign of r when it is written as x + k; use r = −k, not r = k.

Factoring Differences of Squares

Factor the expression 9x2 − 16 completely over the real numbers.

(3x − 4)(3x + 4). Use a2 − b2 = (a − b)(a + b) where a = 3x and b = 4.

Watch out: Writing (3x − 4)2 or (9x − 16)(1), which incorrectly treats it as a perfect square or fails to factor the squares first.

Polynomial Zeros and End Behavior

A polynomial p(x) has degree 5 and a positive leading coefficient. What is the maximum number of real zeros it can have, and how does it behave as x approaches infinity?

It can have at most 5 real zeros. As x approaches infinity, p(x) approaches positive infinity because the end behavior follows the leading term.

Watch out: Confusing the degree with the exact number of zeros; a degree n polynomial has AT MOST n real zeros, not exactly n.

Zeros vs Intercepts Confusion

A student says the zero of f(x) = x2 − 4 is the point (0, −4). What is wrong with this statement and what is correct?

The zero is the input that makes the output 0, not the y-intercept. The zero is x = 2 (and x = −2). The point (0, −4) is the y-intercept.

Watch out: Equating a zero directly with an ordered pair or confusing it with where the graph crosses the y-axis.

Sample questions

Question 1

The expression (x + 1)(x − 2) is equivalent to x² + bx + c, where b and c are constants. What is the value of b?

  1. −2
  2. 3
  3. 1
  4. −1
Show answer and explanation

Answer: −1

Expanding: x² + (1 − 2)x + (1)(−2). So b = −1 and c = −2.

Question 2

The function f(x) = x² + bx + 12, where b is a constant, satisfies f(2) = 0. What is the other zero of f?

  1. −6
  2. 2
  3. 4
  4. 6
Show answer and explanation

Answer: 6

f(2) = 4 + 2b + 12 = 0 → b = −8. Then x² − 8x + 12 = (x − 2)(x − 6), so the other zero is 6. (Shortcut: product of zeros = 12, so other zero = 12/2 = 6.)

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