Algebra and functions

Unit P2: Pure Mathematics 2 - Algebra and Functions

Hello Mathematicians!

Welcome to the Pure Mathematics 2 chapter on Algebra and Functions. In P1, you mastered quadratic expressions and simple simultaneous equations. Now, we are leveling up! We will learn how to handle much more complex polynomial expressions, specifically focusing on division and finding factors efficiently.

Why is this important? These techniques are crucial for solving cubic and quartic equations, which frequently appear in advanced calculus and applied mathematics problems. Master these tools, and you'll find solving tricky equations much easier!

1. Polynomials: Definitions and Algebraic Long Division

1.1 What is a Polynomial?

A polynomial, usually denoted as \(f(x)\) or \(P(x)\), is an expression consisting of variables and coefficients, involving only non-negative integer powers of the variable (like \(x^2, x^3\), but not \(x^{-1}\) or \(\sqrt{x}\)).

• The Degree of the polynomial is the highest power of \(x\).

  Example: \(f(x) = 2x^3 - 5x + 1\) is a polynomial of degree 3 (a cubic).

1.2 Algebraic Long Division

Just like you can divide 17 by 5 to get a quotient of 3 and a remainder of 2, we can divide one polynomial by another. We use Algebraic Long Division when the divisor is not a simple factor, or when we need to find the full quotient.

The Goal: To rewrite the division in the form:

\[ \frac{\text{Dividend}}{\text{Divisor}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}} \]

Which is often written as:

\[ f(x) = Q(x)D(x) + R(x) \]

Where \(Q(x)\) is the Quotient, \(D(x)\) is the Divisor, and \(R(x)\) is the Remainder.

Step-by-Step Process for Long Division

Let's divide \(f(x) = x^3 + 2x^2 - 5x - 6\) by \((x - 2)\).

1. Prepare: Ensure the dividend has no missing powers. If \(x^3 + 5\) was the dividend, you must write it as \(x^3 + 0x^2 + 0x + 5\). (This keeps columns tidy!)
2. Divide the Leading Terms: Divide the first term of the dividend (\(x^3\)) by the first term of the divisor (\(x\)). \(x^3 \div x = x^2\). This is the first term of your quotient.
3. Multiply: Multiply the result (\(x^2\)) by the entire divisor \((x - 2)\). \(x^2(x - 2) = x^3 - 2x^2\).
4. Subtract: Subtract this result from the dividend. BE CAREFUL with signs! \((x^3 + 2x^2) - (x^3 - 2x^2) = 4x^2\).
5. Bring Down: Bring down the next term (\(-5x\)). Your new expression is \(4x^2 - 5x\).
6. Repeat: Go back to step 2 using \(4x^2 - 5x\) as your new dividend fragment, until the remaining term's degree is less than the divisor's degree.

Common Mistake Alert: The most frequent error is subtraction, especially when subtracting negative terms. Always use brackets when performing the subtraction step!

Key Takeaway: Algebraic long division is a robust method to find the exact quotient and remainder, but it can be lengthy. We need faster ways!

2. The Remainder Theorem

Don't panic! You don't always have to do the full long division just to find the remainder. The Remainder Theorem is a massive time-saver.

2.1 The Principle

The Remainder Theorem states that if a polynomial \(f(x)\) is divided by a linear divisor \((x - a)\), the remainder is simply \(f(a)\).

If the divisor is \((ax - b)\), the remainder is \(f\left(\frac{b}{a}\right)\).

How does this work?

Think back to the division formula: \(f(x) = Q(x)D(x) + R\).

If \(D(x) = (x - a)\), then \(f(x) = Q(x)(x - a) + R\).

If we substitute \(x = a\):

\[ f(a) = Q(a)(a - a) + R \]

Since \((a - a) = 0\), the whole quotient term disappears, leaving:

\[ f(a) = R \]

This is extremely powerful!

2.2 Applying the Remainder Theorem (Step-by-Step)

Example: Find the remainder when \(f(x) = x^3 - 3x^2 + 5\) is divided by \((x + 2)\).

1. Identify the Divisor: The divisor is \((x + 2)\).
2. Find the Test Value: Set the divisor to zero: \(x + 2 = 0\), so the test value is \(x = -2\). (Notice, if the divisor is \((x - a)\), we test \(a\). If it's \((x + a)\), we test \(-a\)).
3. Substitute: Calculate \(f(-2)\).

   \[ f(-2) = (-2)^3 - 3(-2)^2 + 5 \]

4. Calculate:

   \[ f(-2) = -8 - 3(4) + 5 \]

   \[ f(-2) = -8 - 12 + 5 = -15 \]

The remainder is -15.

Memory Aid: To find the value you substitute, just ask: "What value of \(x\) would make the divisor equal to zero?"

2.3 Using the Remainder Theorem to Find Unknowns

A common P2 question involves using the remainder to find an unknown coefficient, \(k\).

Example: If \(f(x) = x^3 + kx^2 - 10\) has a remainder of 2 when divided by \((x - 1)\), find \(k\).

We know that \(f(1) = 2\).

\[ f(1) = (1)^3 + k(1)^2 - 10 = 2 \]

\[ 1 + k - 10 = 2 \]

\[ k - 9 = 2 \]

\[ k = 11 \]

Key Takeaway: The Remainder Theorem allows you to calculate the remainder \(R\) without ever performing the full division, simply by evaluating \(f(x)\) at the value that zeros the divisor.

3. The Factor Theorem

The Factor Theorem is the most important application of the Remainder Theorem, as it provides the core method for solving higher-order equations.

3.1 The Principle

A factor is a divisor that divides exactly, meaning the remainder is zero.

The Factor Theorem states that \((x - a)\) is a factor of the polynomial \(f(x)\) if and only if \(f(a) = 0\).

If you divide \(f(x)\) by \((x - a)\) and the remainder is zero, then \(x = a\) is a root (or solution) of the equation \(f(x) = 0\).

3.2 Solving Cubic and Higher Polynomial Equations

To solve \(f(x) = 0\) where \(f(x)\) is a cubic or quartic, you need to find at least one factor using the Factor Theorem. The process is:

1. Test for Factors: Test simple integer values for \(x\) that are factors of the constant term in \(f(x)\). Start with easy values like \(x = \pm 1, \pm 2, \pm 3\).
2. Find One Root: Find a value \(x = a\) such that \(f(a) = 0\). This confirms that \((x - a)\) is a factor.
3. Divide: Use Algebraic Long Division (or inspection) to divide \(f(x)\) by \((x - a)\). This will yield a quadratic quotient \(Q(x)\).

   \[ f(x) = (x - a)Q(x) \]

4. Factor the Quadratic: Solve or factorise the resulting quadratic \(Q(x) = 0\) to find the remaining two roots.

Example Walkthrough: Solve \(x^3 + 4x^2 - 11x - 30 = 0\).

Step 1 & 2 (Testing): The constant term is -30. Factors of 30 are \(\pm 1, \pm 2, \pm 3, \ldots\).

• Try \(x = 1\): \(f(1) = 1 + 4 - 11 - 30 \ne 0\).
• Try \(x = -2\): \(f(-2) = (-8) + 4(4) - 11(-2) - 30 = -8 + 16 + 22 - 30 = 0\).

Since \(f(-2) = 0\), \((x + 2)\) is a factor.

Step 3 (Division): Divide \(x^3 + 4x^2 - 11x - 30\) by \((x + 2)\).

The resulting quotient is \(x^2 + 2x - 15\).

So, \(f(x) = (x + 2)(x^2 + 2x - 15)\).

Step 4 (Factor Quadratic): Factor the quadratic part.

\[ x^2 + 2x - 15 = (x + 5)(x - 3) \]

The equation becomes \((x + 2)(x + 5)(x - 3) = 0\).

The roots are \(x = -2, x = -5\), and \(x = 3\).

Tip for Inspection (Alternative to Long Division):
If you know \((x+2)\) is a factor, you know the quadratic must look like \((x+2)(Ax^2 + Bx + C)\).

By comparing coefficients:

1. The \(x^3\) coefficient must match: \(A = 1\).

2. The constant term must match: \(2 \times C = -30\), so \(C = -15\).

3. Look at the \(x^2\) term: \(4x^2\). This comes from \(2(Ax^2) + Bx(x)\). \(2(1x^2) + Bx^2 = 4x^2\). So \(2 + B = 4\), meaning \(B = 2\).

This gives the quotient \(x^2 + 2x - 15\) instantly!

4. Advanced Application of the Theorems

4.1 The Case of \(ax - b\)

Remember, if the divisor is not monic (i.e., the coefficient of \(x\) is not 1), you must use the fractional form of the substitution.

Example: Find \(k\) if \((2x - 1)\) is a factor of \(f(x) = 4x^3 - 2x^2 + kx - 1\).

1. Set divisor to zero: \(2x - 1 = 0 \implies x = \frac{1}{2}\).
2. Since it's a factor, \(f\left(\frac{1}{2}\right) = 0\).
3. Substitute:

   \[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 2\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) - 1 = 0 \]

   \[ 4\left(\frac{1}{8}\right) - 2\left(\frac{1}{4}\right) + \frac{k}{2} - 1 = 0 \]

   \[ \frac{1}{2} - \frac{1}{2} + \frac{k}{2} - 1 = 0 \]

   \[ \frac{k}{2} - 1 = 0 \implies k = 2 \]

Important Note: When dealing with cubic equations, you must find all three roots. If the quadratic resulting from division cannot be factored, you must use the quadratic formula or completing the square to find the remaining roots (which might be surds or irrational numbers).

Did you know? The general form of a cubic equation was solved completely in the 16th century by Italian mathematicians, long before formal calculus was invented!

Key Takeaway: The Factor Theorem and Remainder Theorem are inseparable tools for manipulating and solving polynomials efficiently. They allow you to transform a high-degree polynomial problem into a simpler quadratic problem.