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Polynomial functions and graphs (QCE Math Methods Unit 1)

Syllabus dot point

“Sketch and analyse polynomial functions of degree 3 and 4, using factored form to read roots and multiplicities, and applying the factor and remainder theorems”

8 min read

Quick answer

Use the factor theorem ((x−a)(x-a) is a factor iff P(a)=0P(a)=0) and rational-roots search to factor a cubic or quartic, then divide out the linear factor. From factored form, read roots and their multiplicity (1 crosses, 2 touches, 3 inflects) and get end behaviour from the leading term.

Jump to a section
  1. What this dot point is asking
  2. Remainder theorem
  3. Factor theorem
  4. Rational roots theorem
  5. Procedure to factor a cubic
  6. Polynomial division
  7. End behaviour
  8. Root multiplicities
  9. Sketching from factored form
  10. Turning points and shape
  11. In one sentence
  12. Exam-style questions

What this dot point is asking

QCAA wants you to sketch polynomial functions of degree 33 and 44, use the factor and remainder theorems to find factors, and read intercepts, end behaviour and root multiplicities from the factored form.

Remainder theorem

If P(x)P(x) is divided by (x−a)(x - a), the remainder is P(a)P(a).

Factor theorem

(x−a)(x - a) is a factor of P(x)P(x) if and only if P(a)=0P(a) = 0.

Rational roots theorem

For a polynomial with integer coefficients, any rational root p/qp/q (in lowest terms) has pp dividing the constant term and qq dividing the leading coefficient.

Procedure to factor a cubic

  1. Identify rational-root candidates.
  2. Test by computing P(a)P(a).
  3. Once a root aa is found, divide PP by (x−a)(x - a).
  4. Factor the resulting quadratic.

Polynomial division

Once a factor (x−a)(x - a) is confirmed, divide P(x)P(x) by it to obtain a quotient of degree one less. Long division works for any divisor, while synthetic division is a fast shortcut for linear divisors. The result expresses P(x)=(x−a)Q(x)P(x) = (x - a)Q(x), after which Q(x)Q(x) (a quadratic, for a cubic) is factored by inspection or the quadratic formula. Equating coefficients is an equivalent method: write the quotient with unknown coefficients, expand, and match each power of xx.

End behaviour

Set by leading term axnax^n:

  • nn even, a>0a > 0: both ends +∞+\infty.
  • nn even, a<0a < 0: both ends −∞-\infty.
  • nn odd, a>0a > 0: left −∞-\infty, right +∞+\infty.
  • nn odd, a<0a < 0: left +∞+\infty, right −∞-\infty.

Root multiplicities

For a factor (x−p)k(x - p)^k:

  • k=1k = 1: crosses the xx-axis.
  • k=2k = 2: touches and turns (double root).
  • k=3k = 3: crosses with a horizontal tangent (point of inflection on axis).

Sketching from factored form

  1. Read roots and multiplicities from the factors.
  2. Find yy-intercept by substituting x=0x = 0.
  3. Determine end behaviour from leading term.
  4. Sketch through the roots with the correct crossing/touching behaviour.

Turning points and shape

A degree-nn polynomial has at most n−1n - 1 turning points, so a cubic has at most two and a quartic at most three. Between consecutive roots the graph must turn at least once, which together with the end behaviour and the crossing or touching behaviour at each root is usually enough to produce a correct shape without calculus. Symmetry can help too: an even polynomial (only even powers) is symmetric about the yy-axis, and an odd polynomial (only odd powers) has rotational symmetry about the origin.

Factoring and sketching a cubic

Sketch y=x3−3x2−x+3y = x^3 - 3x^2 - x + 3, labelling all intercepts and end behaviour.

Step 1: Find an integer root using the rational-roots theorem

The constant term is 33 and the leading coefficient is 11, so the rational-root candidates are ±1\pm 1 and ±3\pm 3. Test x=1x = 1:

1−3−1+3=0.1 - 3 - 1 + 3 = 0.

Since P(1)=0P(1) = 0, the factor theorem tells us (x−1)(x - 1) is a factor.

Step 2: Divide out the known factor

Divide x3−3x2−x+3x^3 - 3x^2 - x + 3 by (x−1)(x - 1) to obtain the quadratic cofactor. Polynomial division gives:

y=(x−1)(x2−2x−3).y = (x - 1)(x^2 - 2x - 3).

Step 3: Factor the remaining quadratic

The quadratic x2−2x−3x^2 - 2x - 3 factors by inspection (numbers that multiply to −3-3 and add to −2-2 are −3-3 and 11):

y=(x−1)(x−3)(x+1).y = (x - 1)(x - 3)(x + 1).

Step 4: Read the roots and their multiplicities

Each linear factor appears once (multiplicity 1), so the graph crosses the xx-axis at all three roots: x=−1x = -1, x=1x = 1, and x=3x = 3. The yy-intercept is y(0)=3y(0) = 3.

Step 5: Determine end behaviour and sketch

The leading term is x3x^3: odd degree with a positive coefficient, so the graph comes from −∞-\infty on the left and rises to +∞+\infty on the right. The curve crosses at x=−1x = -1, turns to a local maximum between −1-1 and 11, crosses at x=1x = 1, turns to a local minimum between 11 and 33, and then crosses at x=3x = 3 and rises.

Final answer: Roots at x=−1,1,3x = -1, 1, 3; yy-intercept (0,3)(0, 3); left end −∞-\infty, right end +∞+\infty.

Common traps
Reading a root with the wrong sign
The factor (x+2)(x + 2) gives the root x=−2x = -2, not +2+2.
Ignoring the rational-roots constraint
Testing x=5x = 5 when 55 does not divide the constant term wastes effort; only divisors of the constant over divisors of the leading coefficient can be rational roots.
Forgetting multiplicities
A factor to the first power crosses the axis, a squared factor touches and turns, and a cubed factor crosses with a horizontal tangent.

In one sentence

Polynomial functions of degree 33 or 44 are factored using the rational-roots theorem to find candidate roots, the factor theorem to verify ((x−a)(x - a) is a factor iff P(a)=0P(a) = 0), and polynomial division to extract the linear factor; sketches read roots and multiplicities from the factored form, yy-intercept from x=0x = 0, and end behaviour from the leading term.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks
Paper 2 (complex familiar). Factorise P(x)=2x3+x2−13x+6P(x) = 2x^3 + x^2 - 13x + 6 completely and determine all roots.
Show worked answer →

Rational-root candidates are divisors of 66 over divisors of 22. Test x=2x = 2: P(2)=16+4−26+6=0P(2) = 16 + 4 - 26 + 6 = 0, so (x−2)(x - 2) is a factor.

Divide: 2x3+x2−13x+6=(x−2)(2x2+5x−3)=(x−2)(2x−1)(x+3)2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) = (x - 2)(2x - 1)(x + 3).

Roots: x=2x = 2, x=12x = \tfrac{1}{2}, x=−3x = -3.

Markers reward the rational-root search, verifying P(2)=0P(2) = 0, the division, and factoring the quadratic.

Practice4 marks
Paper 1 (technique). When P(x)=x3+ax2−7x+2P(x) = x^3 + ax^2 - 7x + 2 is divided by (x−1)(x - 1) the remainder is −3-3. (a) Determine aa. (b) Hence state P(1)P(1) and explain whether (x−1)(x - 1) is a factor of P(x)P(x).
Show worked answer →

(a) By the remainder theorem, P(1)=−3P(1) = -3: 1+a−7+2=−31 + a - 7 + 2 = -3, so a−4=−3a - 4 = -3, giving a=1a = 1.

(b) P(1)=−3≠0P(1) = -3 \neq 0, so by the factor theorem (x−1)(x - 1) is not a factor of P(x)P(x).

Markers reward applying the remainder theorem to find aa and using the factor theorem (a factor requires remainder zero) to justify the conclusion.

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