VCE Mathematical Methods (Year 11) · Unit 2 — Functions & Graphs · 12m read

Polynomial Functions — Cubics and Quartics

Have you ever wondered why some graphs loop and wiggle while others just curve smoothly? Cubic and quartic functions (polynomials of degree 3 and 4) create those fascinating shapes with multiple peaks and valleys. The factor theorem is a clever shortcut that lets you find where a polynomial crosses the x-axis by testing simple values, which makes sketching these complex graphs way easier. Mastering this skill is essential for VCE calculus and helps you understand how engineers model real-world phenomena like motion and optimization.

Learning objective: Sketch polynomial functions and apply the factor theorem.

By the end of this lesson you can:

• I apply the factor theorem to find roots.
• I perform polynomial division.
• I sketch cubics and quartics showing intercepts and behaviour.

Polynomials are rollercoasters with predictable rules. Degree controls how many hills and dips; the leading coefficient decides whether the ride starts up or down.

Polynomials in detail

POLYNOMIAL DEFINITION:
• P(x) = aₙx^n + … + a₁x + a₀.
• n = degree. Leading coefficient = aₙ.

END BEHAVIOUR:
• Degree even & a_n > 0 → both arms up.
• Degree even & a_n < 0 → both arms down.
• Degree odd & a_n > 0 → left down, right up.
• Degree odd & a_n < 0 → left up, right down.

FACTOR THEOREM:
• If P(a) = 0, then (x − a) is a factor of P(x).
• Use to find one root by trial (try ±1, ±2, …, factors of constant term).

REMAINDER THEOREM:
• P(a) = remainder when P(x) is divided by (x − a).

POLYNOMIAL DIVISION:
• Long division or synthetic division by a linear factor.
• Express as P(x) = (x − a) Q(x) + R.
• Once a root is found, divide to reduce degree.

MULTIPLICITY:
• Factor (x − a)^k → graph TOUCHES axis if k even, CROSSES if k odd.
• (x − 2)² → bounces off at x = 2.
• (x − 2)³ → flattens through x = 2.

CUBICS IN STANDARD FORMS:
• y = a(x − h)³ + k (stationary point of inflection at (h, k)).
• y = a(x − p)(x − q)(x − r) (three distinct roots).

QUARTICS:
• Degree 4; up to 4 roots, up to 3 turning points.
• Even-degree polynomial.
• Use substitution u = x² for biquadratics like x⁴ − 5x² + 4.

SKETCHING STEPS:
1. Degree + leading coefficient → end behaviour.
2. y-intercept.
3. x-intercepts (factor theorem / division).
4. Multiplicity of each root.
5. Approximate turning points (later calculus).

SIGN DIAGRAMS:
• Mark roots on x-axis.
• Test sign in each interval.
• Useful for solving P(x) > 0 or P(x) < 0.

Polynomial habits

• Try ±1 first: Factor theorem shortcut.
• Multiplicity: Even touches, odd crosses.
• Degree-root link: n roots (including repeats) in ℂ.

Polynomials

• Factor theorem: P(r) = 0 \iff (x-r) \mid P(x)
• Remainder theorem: P(x) \div (x-r) \,\text{remainder} = P(r)

Quick reference

| Concept | Detail |
|---|---|
| Factor theorem | P(a) = 0 ⇒ (x − a) is a factor |
| Remainder theorem | Remainder of P(x) ÷ (x − a) = P(a) |
| Repeated root | Graph touches axis |
| Triple root | Stationary point of inflection |
| Cubic end behaviour | a > 0: down-left, up-right |

Worked example (Medium)

Factorise P(x) = x³ − 4x² + x + 6.

1. Trial a root. P(−1) = −1 − 4 − 1 + 6 = 0 → (x + 1) is a factor.
2. Divide. x³ − 4x² + x + 6 ÷ (x + 1) = x² − 5x + 6.
3. Factor quadratic. x² − 5x + 6 = (x − 2)(x − 3).

Answer: P(x) = (x + 1)(x − 2)(x − 3). Roots x = −1, 2, 3.

Worked example (Hard)

Solve $x^4 - 5x^2 + 4 = 0$ and sketch the quartic.

1. Substitution. Let $u = x^2$. Then $u^2 - 5u + 4 = 0$.
2. Factor the quadratic in u. $(u - 1)(u - 4) = 0 \Rightarrow u = 1$ or $u = 4$.
3. Back-substitute. $x^2 = 1 \Rightarrow x = \pm 1$. $x^2 = 4 \Rightarrow x = \pm 2$.
4. Sketch features. Degree 4, positive leading coefficient → both arms up. Four roots at $x = -2, -1, 1, 2$. y-intercept $= 4$. Three turning points between the roots.

Answer: $x = -2, -1, 1, 2$. Quartic with four distinct roots and both arms pointing upward.

Common mistakes

• Trying to factorise a cubic without first finding a root via the factor theorem Students attempt to "guess" factors of a degree-3 polynomial without a systematic approach Fix: Always test $P(\pm 1), P(\pm 2), \ldots$ (factors of the constant term) to find a root, then divide.
• Forgetting that a repeated root changes the graph's behaviour at the x-axis Students draw the graph crossing at every root regardless of multiplicity Fix: Even multiplicity → touches and bounces; odd multiplicity → crosses (possibly with a flat inflection).
• Incorrectly determining end behaviour for negative leading coefficients Students assume all cubics go from bottom-left to top-right Fix: Check the sign of the leading coefficient: negative leading coefficient reverses the standard end behaviour.

The factor theorem and polynomial division in context

The Factor Theorem states that $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. This seemingly simple result is the key that unlocks polynomial factorisation beyond quadratics.

For cubics, the strategy is: (1) find one root by trial, (2) divide to get a quadratic quotient, and (3) factorise or apply the quadratic formula to the quotient. This reduces a hard problem to a sequence of easier ones.

Synthetic division is a streamlined version of long division for linear divisors. Set up a row of coefficients, bring down the first, multiply by the root, add down, repeat. It's faster than long division and less error-prone once practised.

For quartics, look for patterns: biquadratics like $x^4 - 5x^2 + 4$ can be solved by substitution $u = x^2$. Others may factor into two quadratics: $x^4 + x^2 - 2 = (x^2 + 2)(x^2 - 1)$.

The Rational Root Theorem narrows down candidates: if $P(x) = a_n x^n + \ldots + a_0$, then any rational root $p/q$ has $p$ dividing $a_0$ and $q$ dividing $a_n$. For monic polynomials ($a_n = 1$), rational roots must be integer factors of the constant term.

VCAA often pairs factorisation with sketching. Once you have the factored form, multiplicity determines the shape at each root, and the leading term determines end behaviour. Combining these gives a reliable sketch without calculus.

Practice questions

1. The remainder when x³ − 2x² + 3 is divided by (x − 2) is:

   1. 0
   2. 3
   3. −1
   4. 7

   (1 marks)

2. Sketch y = (x − 1)²(x + 2). Describe end behaviour and behaviour at each root.

   (3 marks)

Summary

• Apply the factor theorem to find roots.
• Perform polynomial division.
• Sketch cubics and quartics showing intercepts and behaviour.