IP AMaths Notes (Upper Sec, Year 3-4): 09) Polynomials

Polynomials underpin factorisation, curve sketching, and inequality solving. Memorise the theorems that convert substitution into proofs of factor status.

Key results

• Factor theorem: if \( f(a) = 0 \), then \( (x - a) \) divides \( f(x) \).
• Remainder theorem: dividing \( f(x) \) by \( x - a \) leaves remainder \( f(a) \).
• For cubic \( ax^3 + bx^2 + cx + d \), the sum of roots is \( -\tfrac{b}{a} \), pairwise sum \( \tfrac{c}{a} \), product \( -\tfrac{d}{a} \).

Worked example 1 — Factorisation

Given \( f(x) = x^3 - 4x^2 + x + 6 \) and \( f(2) = 0 \), factorise \( f(x) \).

1. Since \( f(2) = 0 \), \( (x - 2) \) is a factor.
2. Perform long division or equate: divide to obtain \( x^2 - 2x - 3 \).
3. Factor quadratic: \( x^2 - 2x - 3 = (x - 3)(x + 1) \).
4. Hence \( f(x) = (x - 2)(x - 3)(x + 1) \).

Worked example 2 — Polynomial inequality

Solve \( x^3 - 4x^2 + x + 6 \leq 0 \).

1. Using factorisation above, inequality is \( (x - 2)(x - 3)(x + 1) \leq 0 \).
2. Critical points: \( x = -1, 2, 3 \).
3. Sign chart:
   • For \( x < -1 \), all factors negative → product negative → included.
   • Between \( -1 \) and \( 2 \), two factors negative, one positive → product positive → excluded.
   • Between \( 2 \) and \( 3 \), one factor negative → product negative → included.
   • Beyond \( 3 \), all positive → product positive → excluded.
4. Include roots because inequality is non-strict.

Solution set: \( x \in (-\infty, -1] \cup [2, 3] \).

Try this

A cubic \( g(x) = 2x^3 + px^2 + qx - 12 \) has factors \( (x - 3) \) and \( (x + 2) \). Find \( p \), \( q \), and the third factor.