Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — factorising a cubic completely
Factorise x3−2x2−5x+6 completely and state its roots.
- 1Test the factors of the constant term 6: ±1,±2,±3,±6.Any rational root must divide the constant term.
- 2f(1)=1−2−5+6=0, so (x−1) is a factor.The factor theorem. Try ±1 first — they are quickest.
- 3Divide: x3−2x2−5x+6=(x−1)(x2−x−6)Long division, or compare coefficients.
- 4Factorise the quadratic: x2−x−6=(x−3)(x+2)Two numbers multiplying to −6 and adding to −1.
- 5f(x)=(x−1)(x−3)(x+2)Three linear factors — 'completely' means going this far.
- 6Roots: x=1, x=3, x=−2Three roots for a cubic, as expected.
Example 2 — the remainder theorem with a non-unit coefficient
Find the remainder when 2x3−x2+4 is divided by 2x−1.
- 1The remainder is f(a), where a makes the divisor zero.Not the divisor's constant term — the value that kills it.
- 22x−1=0 gives x=21.
- 3f(21)=2(81)−41+4Note there is no x term to substitute into.
- 4=41−41+4=4The remainder is 4, so 2x−1 is not a factor.
Example 3 — finding constants in an identity
Find A and B such that 5x+1≡A(x+2)+B(x−1).
- 1An identity holds for every x, so any value may be substituted.This is what makes the shortcut legitimate.
- 2Let x=1 to kill the B term: 6=3AChoose the value that makes a bracket zero.
- 3A=2
- 4Let x=−2 to kill the A term: −9=−3B
- 5B=3
- 6Check: 2(x+2)+3(x−1)=5x+1 ✓Expanding confirms both constants at once.