Polynomial Functions — High School Algebra 2 practice questions

As x → ∞, what happens to f(x) = −2x³ + 5x² − 7?

1. f(x) → −∞
2. f(x) → ∞
3. f(x) → −7
4. f(x) → 0

Answer: A — f(x) → −∞

A) Correct — The leading term −2x³ dominates and becomes very negative. B) Ignores the negative leading coefficient. C) Uses the constant term. D) Polynomials don't level off.

What are the zeros of f(x) = (x − 1)(x + 2)(x − 3)?

1. 1, −2, and 3
2. −1, 2, and −3
3. 1, 2, and 3
4. −6

Answer: A — 1, −2, and 3

A) Correct — Set each factor equal to 0. B) Reverses every sign. C) Ignores the sign in (x + 2). D) Multiplies the constants.

Which is a polynomial function?

1. f(x) = 3x⁴ − x + 2
2. f(x) = 2/x
3. f(x) = √x + 1
4. f(x) = 2ˣ

Answer: A — f(x) = 3x⁴ − x + 2

A) Correct — Polynomials have whole-number exponents. B) Has x in a denominator. C) Has a fractional exponent. D) Is exponential.

Factor x³ + 8.

1. (x + 2)³
2. (x + 2)(x² + 2x + 4)
3. (x + 2)(x² − 2x + 4)
4. (x − 2)(x² + 2x + 4)

Answer: C — (x + 2)(x² − 2x + 4)

C) Correct — a³ + b³ = (a + b)(a² − ab + b²). A) Would expand to x³ + 6x² + 12x + 8. B) Uses the sign pattern for a difference of cubes. D) Is x³ − 8.

What is the end behavior of f(x) = −2x⁴ + 3x² − 1?

1. Both ends go up
2. Down on the left, up on the right
3. Both ends go down
4. Up on the left, down on the right

Answer: C — Both ends go down

C) Correct — Even degree with a negative leading coefficient. A) Would need a positive leading coefficient. B) Describes an odd-degree, positive-leading polynomial. D) Describes an odd-degree, negative-leading polynomial.

Expand (x + 2)³.

1. x³ + 6x² + 12x + 8
2. x³ + 8
3. x³ + 6x² + 6x + 8
4. x³ + 2x² + 4x + 8

Answer: A — x³ + 6x² + 12x + 8

A) Correct — Coefficients 1, 3, 3, 1 with powers of 2. B) Cubes each term separately. C) Forgets the 2² in the third term. D) Leaves out the binomial coefficients.

By the Fundamental Theorem of Algebra, a polynomial of degree n has:

1. exactly n complex roots, counting multiplicity
2. exactly n real roots
3. at most n − 1 roots
4. at least one real root

Answer: A — exactly n complex roots, counting multiplicity

A) Correct — Some of them may be real. B) Some roots may be non-real. C) Confuses roots with turning points. D) x² + 1 has no real roots.

A polynomial of degree 5 can have at most how many turning points?

1. 4
2. 5
3. 6
4. 2

Answer: A — 4

A) Correct — A degree-n polynomial has at most n − 1 turning points. B) Confuses turning points with zeros. C) Adds 1 instead of subtracting. D) Halves the degree.

What is the remainder when p(x) = x³ − 2x² + x − 5 is divided by (x − 1)?

1. 5
2. −9
3. 0
4. −5

Answer: D — −5

D) Correct — By the Remainder Theorem, the remainder is p(1) = 1 − 2 + 1 − 5. A) Gets the sign wrong. B) Evaluates p(−1) instead of p(1). C) Assumes (x − 1) is a factor.

What is the coefficient of x² in (x + 3)⁴?

1. 6
2. 9
3. 18
4. 54

Answer: D — 54

D) Correct — C(4, 2) · 3² = 6 · 9. A) Leaves out 3². B) Leaves out C(4, 2). C) Uses 3 instead of 3².

Worked Example: Use the Rational Root Theorem to find rational roots of p(x) = 2x³ − x² − 7x + 6.

Step 1: List factors of the constant term (6). ±1, ±2, ±3, ±6 Step 2: List factors of the leading coefficient (2). ±1, ±2 Step 3: Form all candidate rational roots p/q. ±1, ±2, ±3, ±6, ±1/2, ±3/2 Step 4: Test candidates by substitution. p(1) = 2 − 1 − 7 + 6 = 0 ✓ → x = 1 is a root. Step 5: Use synthetic division by (x − 1) to factor. 1 | 2 −1 −7 6 2 1 −6 2 1 −6 0 Quotient: 2x² + x − 6 = (2x − 3)(x + 2) Step 6: Solve for remaining roots. 2x − 3 = 0 → x = 3/2 x + 2 = 0 → x = −2 Final answer: rational roots are x = 1, 3/2, and −2.

Multiply: (2x + 3)(x − 4).

2x² − 5x − 12.

Worked Example: Divide (2x³ − 5x² + 3x − 4) by (x − 2) using long division.

Step 1: Set up long division. Write the dividend with all degree terms. 2x³ − 5x² + 3x − 4 ÷ (x − 2) Step 2: Divide leading term 2x³ by x → 2x². Multiply 2x²·(x − 2) = 2x³ − 4x². Subtract. (2x³ − 5x²) − (2x³ − 4x²) = −x² Bring down next: −x² + 3x. Step 3: Divide −x² by x → −x. Multiply −x·(x − 2) = −x² + 2x. Subtract. (−x² + 3x) − (−x² + 2x) = x Bring down next: x − 4. Step 4: Divide x by x → 1. Multiply 1·(x − 2) = x − 2. Subtract. (x − 4) − (x − 2) = −2 Step 5: Write result. Quotient: 2x² − x + 1; Remainder: −2. So (2x³ − 5x² + 3x − 4)/(x − 2) = 2x² − x + 1 − 2/(x − 2) Final answer: quotient 2x² − x + 1, remainder −2.

Factor: x² − 6x + 9.

(x − 3)². Perfect square trinomial.