Welcome to P2 Algebra!
Welcome to the first chapter of Pure Mathematics 2! If you've already completed Paper 1, you have a solid foundation. In this chapter, we are going to add two very powerful tools to your mathematical toolbox: Modulus Functions (handling absolute values) and Advanced Polynomials (dividing them and finding their secrets). Algebra is the "language" of math, and mastering these concepts will make the rest of your AS Level journey much smoother. Don't worry if it seems a bit abstract at first—we'll break it down step-by-step!
1. The Modulus Function \( |x| \)
The modulus of a number is simply its "size" or "magnitude," regardless of whether it is positive or negative. You can think of it as the distance from zero on a number line.
Analogy: Imagine you walk 5 meters to the right (+5) or 5 meters to the left (-5). In both cases, your "distance" is just 5 meters. That distance is the modulus!
• If \( x \) is positive, \( |x| \) is just \( x \). (Example: \( |5| = 5 \))
• If \( x \) is negative, \( |x| \) is the positive version of that number. (Example: \( |-5| = 5 \))
Sketching the Graph of \( y = |ax + b| \)
The graph of a modulus function usually looks like a "V" shape. This is because the "y" value can never be negative.
Step-by-Step Sketching:
1. Imagine the graph without the modulus signs (e.g., \( y = 2x - 4 \)). This is just a straight line.
2. Sketch that line lightly with a pencil.
3. Any part of the line that is below the x-axis (where y is negative) must be reflected (flipped) upward so it becomes positive.
4. The point where the "V" touches the x-axis is called the vertex. You find it by setting the inside of the modulus to zero (e.g., \( 2x - 4 = 0 \), so \( x = 2 \)).
Did you know? The modulus function acts like a "mirror" placed on the x-axis. It ignores anything happening in the negative territory and reflects it back into the positive!
Solving Modulus Equations and Inequalities
When you see \( |a| = |b| \), the easiest way to solve it is to square both sides. This works because squaring any number (positive or negative) always results in a positive value: \( a^2 = b^2 \).
Example: To solve \( |3x - 2| = |2x + 7| \):
1. Square both: \( (3x - 2)^2 = (2x + 7)^2 \)
2. Expand: \( 9x^2 - 12x + 4 = 4x^2 + 28x + 49 \)
3. Move everything to one side and solve the quadratic!
Handling Inequalities:
• For "Less Than" (\( |x - a| < b \)): This means \( x \) is "trapped" between two values. You solve it as: \( a - b < x < a + b \).
• For "Greater Than" (\( |x - a| > b \)): This means \( x \) is "outside" the range. You solve two separate parts: \( x - a > b \) or \( x - a < -b \).
Key Takeaway: The modulus always keeps things positive. When solving equations with modulus on both sides, squaring is your best friend!
2. Polynomial Division
In Paper 1, you worked with quadratics (degree 2). Now, we are moving up to higher powers like \( x^3 \) and \( x^4 \). Sometimes we need to divide these large polynomials by smaller ones (like \( x - 2 \)).
Wait, I thought division was for numbers?
It's exactly the same! Think back to primary school long division. You find how many times the divisor fits, subtract, and find the remainder. We do the same with \( x \)'s.
The Process of Long Division
1. Divide: Look at the first term of the dividend (the big polynomial) and the first term of the divisor. Divide them.
2. Multiply: Multiply your result by the entire divisor.
3. Subtract: Subtract that from your dividend.
4. Bring Down: Bring down the next term and repeat until you can't divide anymore.
The result on top is the Quotient, and what's left at the bottom is the Remainder.
Common Mistake to Avoid: When subtracting, be very careful with negative signs! If you are subtracting \( -5x \), you are actually adding \( 5x \). Most errors in this chapter come from simple sign mistakes.
Key Takeaway: Polynomial division is just "Number Long Division" but with letters. Keep your columns neat (all \( x^2 \) terms in one line, etc.) to avoid confusion.
3. The Remainder and Factor Theorems
What if you don't need the full result of a division and only want to know the remainder? There is a shortcut!
The Remainder Theorem
If you divide a polynomial \( f(x) \) by \( (ax - b) \), the remainder is simply \( f(\frac{b}{a}) \).
In simple terms: If you want to find the remainder when \( f(x) \) is divided by \( (x - 2) \), just plug 2 into the function!
Example: Find the remainder of \( f(x) = x^3 + 5 \) divided by \( (x - 1) \).
Just calculate \( f(1) = (1)^3 + 5 = 6 \). The remainder is 6!
The Factor Theorem
This is a special case of the Remainder Theorem. If you plug a number into the polynomial and the answer is zero, then that expression is a factor (there is no remainder).
If \( f(c) = 0 \), then \( (x - c) \) is a factor.
How to use this to solve equations:
1. If you have a cubic equation like \( x^3 - 6x^2 + 11x - 6 = 0 \), try plugging in small numbers like 1, -1, 2, -2.
2. If \( f(1) = 0 \), you know \( (x - 1) \) is a factor.
3. Use long division to divide the cubic by \( (x - 1) \).
4. You'll be left with a quadratic, which you can solve easily using the formula or factorising!
Quick Review Box:
• Remainder Theorem: Plug the value in \(\rightarrow\) Get the remainder.
• Factor Theorem: Plug the value in \(\rightarrow\) Get zero \(\rightarrow\) You found a factor!
• Solving: Factor Theorem + Long Division = Solving complex equations.
Key Takeaway: Don't waste time with long division if the question only asks for the remainder. Use the theorem! It’s faster and has fewer places to make mistakes.
Summary Checklist
Don't worry if this seems tricky at first! Algebra is about practice. Before you move on, make sure you can:
• Recognize the "V" shape of a modulus graph.
• Solve modulus equations by squaring both sides.
• Perform long division without losing track of your negative signs.
• Use the Factor Theorem to "break down" a cubic equation into a quadratic.
Keep going! You've just mastered the foundational algebra for Pure Math 2.