Polynomial functions and graphs (QCE Math Methods Unit 1)
“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”
Use the factor theorem ( is a factor iff ) 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.
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What this dot point is asking
QCAA wants you to sketch polynomial functions of degree and , use the factor and remainder theorems to find factors, and read intercepts, end behaviour and root multiplicities from the factored form.
Remainder theorem
If is divided by , the remainder is .
Factor theorem
is a factor of if and only if .
Rational roots theorem
For a polynomial with integer coefficients, any rational root (in lowest terms) has dividing the constant term and dividing the leading coefficient.
Procedure to factor a cubic
- Identify rational-root candidates.
- Test by computing .
- Once a root is found, divide by .
- Factor the resulting quadratic.
Polynomial division
Once a factor is confirmed, divide 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 , after which (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 .
End behaviour
Set by leading term :
- even, : both ends .
- even, : both ends .
- odd, : left , right .
- odd, : left , right .
Root multiplicities
For a factor :
- : crosses the -axis.
- : touches and turns (double root).
- : crosses with a horizontal tangent (point of inflection on axis).
Sketching from factored form
- Read roots and multiplicities from the factors.
- Find -intercept by substituting .
- Determine end behaviour from leading term.
- Sketch through the roots with the correct crossing/touching behaviour.
Turning points and shape
A degree- polynomial has at most 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 -axis, and an odd polynomial (only odd powers) has rotational symmetry about the origin.
Sketch , labelling all intercepts and end behaviour.
Step 1: Find an integer root using the rational-roots theorem
The constant term is and the leading coefficient is , so the rational-root candidates are and . Test :
Since , the factor theorem tells us is a factor.
Step 2: Divide out the known factor
Divide by to obtain the quadratic cofactor. Polynomial division gives:
Step 3: Factor the remaining quadratic
The quadratic factors by inspection (numbers that multiply to and add to are and ):
Step 4: Read the roots and their multiplicities
Each linear factor appears once (multiplicity 1), so the graph crosses the -axis at all three roots: , , and . The -intercept is .
Step 5: Determine end behaviour and sketch
The leading term is : odd degree with a positive coefficient, so the graph comes from on the left and rises to on the right. The curve crosses at , turns to a local maximum between and , crosses at , turns to a local minimum between and , and then crosses at and rises.
Final answer: Roots at ; -intercept ; left end , right end .
- Reading a root with the wrong sign
- The factor gives the root , not .
- Ignoring the rational-roots constraint
- Testing when 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 or are factored using the rational-roots theorem to find candidate roots, the factor theorem to verify ( is a factor iff ), and polynomial division to extract the linear factor; sketches read roots and multiplicities from the factored form, -intercept from , 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 marksPaper 2 (complex familiar). Factorise completely and determine all roots.Show worked answer →
Rational-root candidates are divisors of over divisors of . Test : , so is a factor.
Divide: .
Roots: , , .
Markers reward the rational-root search, verifying , the division, and factoring the quadratic.
Practice4 marksPaper 1 (technique). When is divided by the remainder is . (a) Determine . (b) Hence state and explain whether is a factor of .Show worked answer →
(a) By the remainder theorem, : , so , giving .
(b) , so by the factor theorem is not a factor of .
Markers reward applying the remainder theorem to find and using the factor theorem (a factor requires remainder zero) to justify the conclusion.