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NSWMaths Extension 1Syllabus dot point

How do we solve polynomial and rational inequalities using sign analysis?

Solve polynomial and rational inequalities by factoring and analysing the sign of each factor across critical values

A focused answer to the HSC Maths Extension 1 dot point on solving polynomial and rational inequalities. Sign tables, critical values, multiplicity behaviour, and the special care needed when a denominator is involved, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answer

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What this dot point is asking

NESA wants you to solve polynomial inequalities like P(x)β‰₯0P(x) \ge 0 and rational inequalities like P(x)Q(x)<0\frac{P(x)}{Q(x)} < 0 by analysing the sign of the expression across the intervals carved out by its critical values.

The answer

The sign-table method

  1. Factor the polynomial or rational expression completely.
  2. Identify critical values: where each factor equals zero (sign-change candidates) and, for rational expressions, where the denominator is zero (excluded values).
  3. Order the critical values on a number line.
  4. Determine the sign of each factor in each interval.
  5. Multiply the signs to get the sign of the whole expression.
  6. Combine intervals based on the direction of the inequality and the type of endpoint (open or closed).

Critical values

A polynomial P(x)P(x) changes sign across each root of odd multiplicity, and does not change sign across roots of even multiplicity.

A rational expression P(x)Q(x)\frac{P(x)}{Q(x)} has critical values at the zeros of both PP and QQ. Zeros of QQ are excluded from the solution because the expression is undefined there.

Handling multiplicity

A factor like (xβˆ’2)2(x - 2)^2 touches the x-axis but does not cross. The sign is the same either side. Treat it as not changing the sign of the product.

A factor like (xβˆ’2)3(x - 2)^3 behaves like (xβˆ’2)(x - 2) for sign purposes: sign changes across x=2x = 2.

The shortcut: for sign analysis, count multiplicities modulo 22. Even multiplicities can be ignored.

Rational inequalities: never multiply by an expression of unknown sign

A common rookie error is to multiply both sides of ABβ‰₯0\frac{A}{B} \ge 0 by BB to clear the denominator. This is invalid because BB might be negative. Instead, move everything to one side and analyse the sign of the rational expression as a single object.

If you must clear the denominator, multiply by B2B^2 (always non-negative) instead. Or split into cases based on the sign of BB. The sign-table method avoids these pitfalls.

Quadratic inequalities

For ax2+bx+cβ‰₯0a x^2 + b x + c \ge 0 with a>0a > 0, the parabola opens upward.

  • If Ξ”>0\Delta > 0 (two real roots Ξ±<Ξ²\alpha < \beta): expression is positive outside the roots, that is x≀αx \le \alpha or xβ‰₯Ξ²x \ge \beta.
  • If Ξ”=0\Delta = 0: expression is β‰₯0\ge 0 everywhere.
  • If Ξ”<0\Delta < 0: expression is positive everywhere.

For a<0a < 0, all signs flip.

Endpoint conventions

  • Strict inequality (>> or <<): exclude critical values where the expression equals zero. Always exclude where the expression is undefined.
  • Weak inequality (β‰₯\ge or ≀\le): include zeros of the numerator, exclude zeros of the denominator (still undefined).

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q63 marksSolve the inequality xβˆ’1x+2β‰₯0\frac{x - 1}{x + 2} \ge 0.
Show worked answer β†’

Critical values: numerator zero at x=1x = 1, denominator zero at x=βˆ’2x = -2 (excluded).

Sign table across (βˆ’βˆž,βˆ’2)(-\infty, -2), (βˆ’2,1)(-2, 1) and (1,∞)(1, \infty):

  • IMATH_5 : numerator negative, denominator negative, quotient positive.
  • IMATH_6 : numerator negative, denominator positive, quotient negative.
  • IMATH_7 : numerator positive, denominator positive, quotient positive.

The inequality is β‰₯0\ge 0, so include the value where the expression equals zero (x=1x = 1) but exclude where it is undefined (x=βˆ’2x = -2).

Solution: x<βˆ’2x < -2 or xβ‰₯1x \ge 1, that is x∈(βˆ’βˆž,βˆ’2)βˆͺ[1,∞)x \in (-\infty, -2) \cup [1, \infty).

Markers reward identifying critical values, building a sign table, handling the strict-versus-equality endpoints correctly, and excluding the undefined value.

2020 HSC Q83 marksFind all real values of xx for which (xβˆ’1)(x+3)(xβˆ’2)>0(x - 1)(x + 3)(x - 2) > 0.
Show worked answer β†’

Critical values: x=βˆ’3,1,2x = -3, 1, 2.

Build a sign table across (βˆ’βˆž,βˆ’3)(-\infty, -3), (βˆ’3,1)(-3, 1), (1,2)(1, 2), (2,∞)(2, \infty):

  • IMATH_5 : (βˆ’)(βˆ’)(βˆ’)=βˆ’(-)(-)(-) = -.
  • IMATH_7 : (βˆ’)(+)(βˆ’)=+(-)(+)(-) = +.
  • IMATH_9 : (+)(+)(βˆ’)=βˆ’(+)(+)(-) = -.
  • IMATH_11 : (+)(+)(+)=+(+)(+)(+) = +.

The inequality is strict (>>), so exclude the critical values.

Solution: βˆ’3<x<1-3 < x < 1 or x>2x > 2, that is x∈(βˆ’3,1)βˆͺ(2,∞)x \in (-3, 1) \cup (2, \infty).

Markers reward critical values, sign analysis (alternation in this case because all roots have multiplicity 1), and correct interval notation.

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