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QLDMath MethodsSyllabus dot point

How are polynomial functions analysed?

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

A focused answer to the QCE Math Methods Unit 1 dot point on polynomial functions. Sketches cubics and quartics from factored form, applies the factor and remainder theorems, and works the standard QCAA factor-a-cubic problem.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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 (xβˆ’a)(x - a), the remainder is P(a)P(a).

Factor theorem

(xβˆ’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 (xβˆ’a)(x - a).
  4. Factor the resulting quadratic.

Polynomial division

Once a factor (xβˆ’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)=(xβˆ’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 βˆ’βˆž-\infty.
  • nn odd, a>0a > 0: left βˆ’βˆž-\infty, right +∞+\infty.
  • nn odd, a<0a < 0: left +∞+\infty, right βˆ’βˆž-\infty.

Root multiplicities

For a factor (xβˆ’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 nβˆ’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.

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 ((xβˆ’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 practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20225 marksPaper 2 (complex familiar). Factorise P(x)=2x3+x2βˆ’13x+6P(x) = 2x^3 + x^2 - 13x + 6 completely and determine all roots.
Show worked answer β†’

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

Divide: 2x3+x2βˆ’13x+6=(xβˆ’2)(2x2+5xβˆ’3)=(xβˆ’2)(2xβˆ’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=βˆ’3x = -3.

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

QCAA 20234 marksPaper 1 (technique). When P(x)=x3+ax2βˆ’7x+2P(x) = x^3 + ax^2 - 7x + 2 is divided by (xβˆ’1)(x - 1) the remainder is βˆ’3-3. (a) Determine aa. (b) Hence state P(1)P(1) and explain whether (xβˆ’1)(x - 1) is a factor of P(x)P(x).
Show worked answer β†’

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

(b) P(1)=βˆ’3β‰ 0P(1) = -3 \neq 0, so by the factor theorem (xβˆ’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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