Polynomial Functions (30-1)

Question 1

Define a polynomial function.

Answer 1

A function that can be written as f(x) = a_n x^n + a_(n-1) x^(n-1) + … + a_1 x + a_0 where n is a whole number and a_n ≠ 0.

Question 2

What is the degree of a polynomial?

Answer 2

The highest exponent of x in the polynomial.

Question 3

What does the leading coefficient tell you?

Answer 3

Combined with degree, it determines end behaviour (up/down direction of the graph’s ends).

Question 4

Describe the end behaviour of an even-degree polynomial with positive leading coefficient.

Answer 4

Both ends rise (up-up).

Question 5

Describe the end behaviour of an even-degree polynomial with negative leading coefficient.

Answer 5

Both ends fall (down-down).

Question 6

Describe the end behaviour of an odd-degree polynomial with positive leading coefficient.

Answer 6

Left end down, right end up.

Question 7

Describe the end behaviour of an odd-degree polynomial with negative leading coefficient.

Answer 7

Left end up, right end down.

Question 8

What is the Factor Theorem?

Answer 8

If f(c) = 0, then (x - c) is a factor of f(x).

Question 9

What is the Remainder Theorem?

Answer 9

The remainder when f(x) is divided by (x - c) is f(c).

Question 10

Find the remainder when f(x) = x^3 - 4x + 1 is divided by x - 2.

Answer 10

Use Remainder Theorem: f(2) = 2^3 - 4(2) + 1 = 8 - 8 + 1 = 1 → Remainder = 1.

Question 11

Explain multiplicity of a root.

Answer 11

Multiplicity is the number of times a root is repeated. Odd multiplicity crosses the x-axis, even multiplicity just touches (bounces).

Question 12

Give an example of a polynomial with a double root at x = 3 and a single root at x = -1.

Answer 12

f(x) = a(x - 3)^2(x + 1).

Question 13

Describe the graph near a triple root.

Answer 13

Crosses the x-axis but flattens (inflection) near the root.

Question 14

Explain synthetic (or long) division.

Answer 14

A method of dividing a polynomial by a linear binomial to find quotient and remainder.

Question 15

Describe how to find possible rational roots using the Rational Root Theorem.

Answer 15

Possible roots = ±(factors of constant term) ÷ (factors of leading coefficient).

Question 16

Sketch the graph of f(x) = (x - 1)(x + 2)(x - 4).

Answer 16

Roots at x = 1, -2, 4; degree 3 so end behaviour down-left, up-right; crosses x-axis at each root.

Question 17

Determine the equation of a cubic passing through (0, -8) with roots -1, 2, 4.

Answer 17

f(x) = a(x+1)(x-2)(x-4). Substitute (0,-8): -8 = a(1)(-2)(-4) → -8 = 8a → a = -1. So f(x) = -(x+1)(x-2)(x-4).

Question 18

Describe how the y-intercept of a polynomial can be found.

Answer 18

Substitute x = 0 into f(x).

Question 19

What is the maximum number of turning points a polynomial of degree n can have?

Answer 19

At most n - 1 turning points.

Question 20

Describe how transformations affect polynomial functions.

Answer 20

Vertical stretch/compression by factor a, reflection over x-axis if a < 0, horizontal/vertical shifts by h and k.

Question 21

Give an example of a quartic function with all real roots.

Answer 21

f(x) = (x-1)(x-2)(x+3)(x+4).

Question 22

Give an example of a quartic function with no real roots.

Answer 22

f(x) = x^4 + 4x^2 + 4 (discriminant < 0 for quadratic in x^2).

Question 23

State how to write a polynomial in mapping notation.

Answer 23

x → a(x - r_1)^(m_1)(x - r_2)^(m_2)… where r_i are roots and m_i are multiplicities.

Question 24

Describe how polynomial graphs behave as x → ±∞ based on degree and leading coefficient.

Answer 24

Even degree → ends in same direction; odd degree → ends in opposite directions; sign of leading coefficient controls up/down.

Question 25

If f(x) = x^4 - 3x^3 + 2x - 5, what is f(-x)?

Answer 25

f(-x) = (-x)^4 - 3(-x)^3 + 2(-x) - 5 = x^4 + 3x^3 - 2x - 5.

Question 26

Explain why a polynomial function is continuous and smooth.

Answer 26

No breaks, holes, or sharp corners; all polynomials are differentiable everywhere.

Question 27

Use synthetic division to divide f(x) = x^3 - 6x^2 + 11x - 6 by (x - 1).

Answer 27

Setup with root 1: Bring down 1, multiply-add steps: 1 | 1 -6 11 -6 → 1 -5 6 0 → Quotient = x^2 - 5x + 6, Remainder = 0.

Question 28

Use synthetic division to divide f(x) = 2x^3 + 3x^2 - 11x - 6 by (x + 2).

Answer 28

Root is -2: Bring down 2, steps → 2 | 2 3 -11 -6 → 2 -1 -9 12 → Quotient = 2x^2 - x - 9, Remainder = 12.

Question 29

What does a remainder of 0 mean after synthetic division?

Answer 29

It means the divisor (x - c) is a factor of the polynomial.

Question 30

Determine if x - 4 is a factor of f(x) = x^3 - 5x^2 + 2x + 8.

Answer 30

Use synthetic division (root 4): 1 -5 2 8 → Bring down 1, 4, -1, 4, 12 → Remainder 12 ≠ 0, so x - 4 is NOT a factor.

Question 31

From the graph of a polynomial, you see roots at x = -3, x = 0, x = 4 (with bounce at 4). Determine the factored form.

Answer 31

f(x) = a(x + 3)(x)(x - 4)^2 (a determined by y-intercept or given point).

Question 32

A cubic polynomial has roots x = -2, x = 1, and passes through (0, -12). Find the equation if leading coefficient a = 2.

Answer 32

f(x) = 2(x + 2)(x - 1)(x - r). Use given point to solve r. Substitute x=0: -12 = 2(2)(-1)(-r) → -12 = 4(–r)(-1)= 4r → r = -3 → f(x) = 2(x+2)(x-1)(x+3).

Question 33

The graph of a quartic has end behaviour up-up, roots at x = -1 (single root), x = 2 (triple root). Write the general equation.

Answer 33

f(x) = a(x + 1)(x - 2)^3 where a > 0.

Question 34

How do you determine the sign of the leading coefficient from a graph?

Answer 34

Look at end behaviour: Up-up = positive leading coefficient, Down-down = negative leading coefficient (for even degree).

Question 35

A graph shows x-intercepts at -2, 0, and 3, with bounce at -2 and flattening at 3. Degree is 4. Find multiplicities.

Answer 35

(-2) has multiplicity 2 (bounce), (0) multiplicity 1 (cross), (3) multiplicity 3 would exceed degree, so must be multiplicity 1 but with a flatter curve → likely a > 0 but needs scaling.

Question 36

Given a graph with known y-intercept (0, 12) and roots -1, 2, 5, construct polynomial.

Answer 36

f(x) = a(x+1)(x-2)(x-5). Substitute x=0, y=12 to solve for a: 12 = a(1)(-2)(-5) = 10a → a = 12/10 = 1.2 → f(x) = 1.2(x+1)(x-2)(x-5).