Polynomial Functions Worksheet Answer Key – Algebra 2

Problem 1 Degree and leading coefficient

Write each in standard form first.

a) degree 4, leading coefficient 5: quartic.

b) degree 3, leading coefficient −3: cubic.

c) degree 5, leading coefficient 6: quintic.

d) degree 6, leading coefficient 4: degree 6.

Problem 2 End behavior

a) Even degree, negative leading coefficient: falls to the left and falls to the right.

b) Odd degree, positive leading coefficient: falls to the left and rises to the right.

c) Odd degree, negative leading coefficient: rises to the left and falls to the right.

d) Even degree, positive leading coefficient: rises to the left and rises to the right.

Problem 3 Zeros from factored form

a) x − 4 = 0, x + 7 = 0, 2x − 5 = 0: 4, −7, 52.

b) x² = 0 or x + 3 = 0: 0 and −3.

Problem 4 Multiplicity

Zero −7: multiplicity 3 (odd) crosses. Zero 3: multiplicity 2 (even) touches. Zero 8: multiplicity 1 crosses.

Degree = 3 + 2 + 1 = 6.

−7 (3, crosses), 3 (2, touches), 8 (1, crosses); degree 6

Problem 5 Reading a polynomial graph

The graph crosses the x-axis at −2, 1, and 3. It has a local maximum between −2 and 1 and a local minimum between 1 and 3: 2 turning points.

With 3 zeros and 2 turning points, the least degree is 3.

Zeros −2, 1, 3; 2 turning points; degree 3

Problem 6 Factoring out a common factor

a) x(x² − 9) = x(x − 3)(x + 3): 0, 3, −3.

b) x(x² + 4x − 21) = x(x + 7)(x − 3): 0, −7, 3.

Problem 7 Quadratic form

Factor like a quadratic in x²: (x² − 1)(x² − 16) = (x − 1)(x + 1)(x − 4)(x + 4).

x = ±1 and x = ±4

Problem 8 Factoring by grouping

Group: x²(x + 2) − 25(x + 2) = (x + 2)(x² − 25) = (x + 2)(x − 5)(x + 5).

−2, 5, −5

Problem 9 Writing a polynomial from its zeros

f(x) = (x + 3)(x − 2)(x − 6). First, (x + 3)(x − 2) = x² + x − 6.

Then (x² + x − 6)(x − 6) = x³ − 6x² + x² − 6x − 6x + 36.

f(x) = x³ − 5x² − 12x + 36

Problem 10 Using a point to find the leading coefficient

f(x) = a(x + 1)²(x − 4). Then f(0) = a(1)(−4) = −4a = −12, so a = 3.

f(x) = 3(x + 1)²(x − 4)

Problem 11 Sketching from factored form

Degree 3, leading coefficient 1: falls left, rises right. Zeros: −3 (crosses), 1 (touches). y-intercept: f(0) = 3.

Crosses at −3, touches at 1, y-intercept 3

Problem 12 Sign of a polynomial

Test one point per interval. f(−5) = (−1)(−6)(−10) = −60. f(0) = (4)(−1)(−5) = 20. f(2) = (6)(1)(−3) = −18. f(6) = (10)(5)(1) = 50.

Negative, positive, negative, positive

Problem 13 Turning points and zeros

Turning points: at most 6 − 1 = 5. Real zeros: at most 6. Yes: f(x) = x⁶ + 1 is always at least 1, so it has no real zeros.

5 turning points; 6 zeros; yes, x⁶ + 1

Problem 14 Where is the function positive?

Zeros: 0 (multiplicity 2), 3 (multiplicity 1), −6 (multiplicity 1). Degree 4, leading coefficient −1: both ends fall.

x² ≥ 0, so the sign comes from −(x − 3)(x + 6), which is positive when (x − 3)(x + 6) < 0, that is, −6 < x < 3. At x = 0 the function equals 0.

f(x) > 0 for −6 < x < 0 and 0 < x < 3

Problem 15 Finding all zeros of a cubic

f(−2) = 2(−8) − 3(4) − 11(−2) + 6 = −16 − 12 + 22 + 6 = 0, so −2 is a zero and (x + 2) is a factor.

Dividing gives f(x) = (x + 2)(2x² − 7x + 3) = (x + 2)(2x − 1)(x − 3).

Zeros −2, 12, 3

Problem 16 An open box from a sheet

The box has height x, width 12 − 2x, and length 18 − 2x, so V(x) = x(12 − 2x)(18 − 2x) = 4x³ − 60x² + 216x.

Each cut must leave positive width: 12 − 2x > 0 and x > 0, so the domain is 0 < x < 6.

Domain 0 < x < 6; greatest volume about 228 cubic inches at x ≈ 2.35 inches

Problem 17 A tunnel in a roller coaster

Zeros: 0, 5, 9. With positive leading coefficient: positive for 0 < x < 5, negative for 5 < x < 9, positive for x > 9.

h(7) = 0.2(7)(2)(−2) = −5.6.

Zeros 0, 5, 9; below ground for 5 < x < 9; height −5.6 feet

Problem 18 A cubic through a point

f(x) = a(x + 2)(x − 1)(x − 4). Then f(2) = a(4)(1)(−2) = −8a = −16, so a = 2.

2(x + 2)(x − 1)(x − 4) = 2(x³ − 3x² − 6x + 8).

f(x) = 2(x + 2)(x − 1)(x − 4) = 2x³ − 6x² − 12x + 16

Problem 19 Identifying a function from its graph

Four zeros (−3, −1, 1, 2), each crossing, so degree at least 4. Both ends fall, so the leading coefficient is negative.

f(x) = a(x + 3)(x + 1)(x − 1)(x − 2). Then f(0) = a(3)(1)(−1)(−2) = 6a = −3, so a = –12.

Degree 4; negative leading coefficient; f(x) = –12(x + 3)(x + 1)(x − 1)(x − 2)

Problem 20 Analyzing a polynomial without graphing

Degree: 1 + 2 + 3 = 6. Leading coefficient: 2.

Even degree and positive leading coefficient: both ends rise.

Zero 1 (multiplicity 1) crosses; zero −2 (multiplicity 2) touches; zero 5 (multiplicity 3) crosses and flattens.

g(0) = 2(−1)(4)(−125) = 1,000.

Degree 6, leading coefficient 2, both ends rise, y-intercept 1,000