Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — a remainder without doing the division
Find the remainder when f(x)=2x3−3x2+x−5 is divided by (x−2).
- 1The divisor (x−2) is zero at x=2.Ask what value kills the bracket. That is the value you substitute.
- 2f(2)=2(8)−3(4)+2−5Substitute carefully, one term at a time.
- 3=16−12+2−5=1The remainder is 1. No long division was needed.
Example 2 — factorising a cubic completely
Factorise f(x)=x3−4x2+x+6 completely, and hence solve f(x)=0.
- 1Try x=−1: (−1)−4(1)+(−1)+6=0Only test factors of the constant term 6: ±1,±2,±3,±6.
- 2So (x+1) is a factor.f(−1)=0, so by the factor theorem (x+1) divides f(x) exactly.
- 3Dividing: x3−4x2+x+6=(x+1)(x2−5x+6)The quotient of a cubic by a linear factor is always a quadratic.
- 4x2−5x+6=(x−2)(x−3)Factorise the quadratic. Do not stop before this — the question said completely.
- 5f(x)=(x+1)(x−2)(x−3), so x=−1,2,3Check: 1×(−2)×(−3)=6, matching the constant term.
Example 3 — two unknowns, two factors
Both (x−1) and (x+2) are factors of f(x)=x3+ax2+bx−6. Find a and b.
- 1f(1)=0: 1+a+b−6=0⇒a+b=5Each factor gives you one equation. This is the first.
- 2f(−2)=0: −8+4a−2b−6=0⇒4a−2b=14Watch the signs: (−2)3=−8 and (−2)2=+4.
- 3Simplify to 2a−b=7, then add to a+b=5Divide the second equation by 2 first — it makes the elimination clean.
- 43a=12⇒a=4, and b=1Substitute back into the simpler equation.
- 5Check: f(x)=x3+4x2+x−6, and f(1)=0, f(−2)=0.Always verify both conditions. In fact f(x)=(x−1)(x+2)(x+3).