Welcome to P2: Algebra and Functions!
In Pure Mathematics 1 (P1), you mastered the art of quadratics. Now, in Pure Mathematics 2 (P2), we are taking it a step further. We are moving beyond the power of 2 and diving into the world of polynomials (like cubics and quartics). In this chapter, you will learn how to divide complex algebraic expressions as easily as you divide numbers, and you'll discover two "magic" theorems that will save you heaps of time in your exams.
Don't worry if this seems a bit more complex than P1 at first—once you see the patterns, it becomes a very logical and satisfying puzzle to solve!
1. Simple Algebraic Division
Think back to primary school when you learned "long division" for numbers. Algebraic division works exactly the same way! We use it to divide a large polynomial (the dividend) by a smaller linear expression (the divisor).
Key Terms to Know:
- Dividend: The thing you are dividing (e.g., in \(13 \div 4\), 13 is the dividend).
- Divisor: The thing you are dividing by (e.g., 4 is the divisor).
- Quotient: The "main" answer you get.
- Remainder: What is left over if it doesn't fit perfectly.
How to do it (Step-by-Step):
To divide a polynomial like \(x^3 + 3x^2 - 4\) by \((x - 1)\):
- Divide: Divide the first term of the dividend (\(x^3\)) by the first term of the divisor (\(x\)). Write this above the line.
- Multiply: Multiply your new term by the whole divisor.
- Subtract: Subtract this result from your original expression to see what's left.
- Bring Down: Bring down the next term and repeat until you reach a constant.
Quick Review: If you divide a polynomial \(f(x)\) and the remainder is 0, then the divisor is a perfect fit!
2. The Remainder Theorem
What if you only need to know the remainder and don't care about the quotient? Long division takes a long time. The Remainder Theorem is your shortcut!
The Rule: When a polynomial \(f(x)\) is divided by \((ax - b)\), the remainder is simply \(f(\frac{b}{a})\).
Example:
Find the remainder when \(f(x) = x^3 - 2x^2 + 4\) is divided by \((x - 2)\).
Instead of doing long division, just plug in \(x = 2\):
\(f(2) = (2)^3 - 2(2)^2 + 4\)
\(f(2) = 8 - 8 + 4 = 4\)
The remainder is 4. Easy!
Memory Aid: Think of the divisor as a "root in disguise." If the divisor is \((x - 3)\), plug in positive 3. If the divisor is \((x + 5)\), plug in negative 5. Always switch the sign!
3. The Factor Theorem
The Factor Theorem is just a special version of the Remainder Theorem. It is one of the most important tools for solving cubic equations.
The Logic: If you divide a number and the remainder is 0, that number is a factor (e.g., \(10 \div 2 = 5\) with 0 remainder, so 2 is a factor of 10).
In algebra: If \(f(\frac{b}{a}) = 0\), then \((ax - b)\) is a factor of \(f(x)\).
Did you know? This is how we "unlock" cubic graphs. If you find a factor \((x - 2)\), you know the graph crosses the x-axis at \(x = 2\)!
4. Factorising Cubic Expressions
The syllabus requires you to factorise expressions like \(x^3 + 3x^2 - 4\). Here is the standard "Battle Plan":
Step 1: The Guess (Trial and Error)
Use the Factor Theorem to find one number that makes the expression equal 0. Start with small numbers like \(1, -1, 2, -2\).
Example: If \(f(1) = 0\), then \((x - 1)\) is your first factor.
Step 2: The Division
Divide your cubic expression by the factor you just found using long division. The answer (quotient) will be a quadratic.
Step 3: Finish it Off
Factorise the quadratic you got in Step 2 (using the methods you learned in P1). Now you have all three factors!
Key Takeaway: A cubic expression can usually be broken down into three linear factors: \((x - p)(x - q)(x - r)\).
5. Common Mistakes to Avoid
- Sign Errors: This is the #1 mistake! When subtracting during long division, remember that subtracting a negative makes a positive.
- Missing Terms: If a polynomial is missing a power (e.g., \(x^3 - 4\)), rewrite it as \(x^3 + 0x^2 + 0x - 4\) before dividing. If you don't keep the columns straight, the division will fail!
- Wrong Substitution: If dividing by \((2x - 1)\), you must substitute \(x = \frac{1}{2}\) into the Remainder Theorem, not \(1\) or \(-1\).
Chapter Summary
- Use algebraic long division to find both the quotient and the remainder.
- Use the Remainder Theorem \(f(\frac{b}{a})\) to find the remainder instantly.
- If \(f(\text{value}) = 0\), the corresponding bracket is a Factor.
- To factorise a cubic: Find one factor, divide to get a quadratic, then factorise the quadratic.
Keep practicing! Algebraic division is a "muscle memory" skill. The more you do it, the more natural it feels. You've got this!