Introduction to Factor and Remainder Theorems

Welcome! In the previous chapter, you learned how to perform simple algebraic division. While long division is a reliable tool, it can be time-consuming. What if you only need to know the remainder? Or what if you want to check if a bracket like \((x - 3)\) is a perfect factor of a complicated cubic expression?

In this chapter, we explore two powerful shortcuts: the Remainder Theorem and the Factor Theorem. These tools allow us to "skip" the long division and get straight to the answers we need. They are essential for solving higher-level equations and factorising cubics in the "Identities and inequalities" section of your syllabus.


1. The Remainder Theorem

Imagine you have a function \(f(x)\) and you divide it by a linear expression like \((x - a)\). Instead of doing the full division to find out what is left over, the Remainder Theorem gives us a brilliant shortcut.

The Rule: When a polynomial \(f(x)\) is divided by \((x - a)\), the remainder is simply \(f(a)\).

If you are dividing by something slightly more complex, like \((ax - b)\), the remainder is \(f(\frac{b}{a})\). Essentially, you take the value of \(x\) that makes the divisor equal to zero and plug it into the function.

Step-by-Step Example:

Find the remainder when \(f(x) = 2x^3 - 5x^2 + 3x - 4\) is divided by \((x - 2)\).

Step 1: Identify the value to substitute. Since we are dividing by \((x - 2)\), we set \(x - 2 = 0\), which gives us \(x = 2\).
Step 2: Substitute this value into the function.
\(f(2) = 2(2)^3 - 5(2)^2 + 3(2) - 4\)
Step 3: Calculate the result.
\(f(2) = 2(8) - 5(4) + 6 - 4\)
\(f(2) = 16 - 20 + 6 - 4 = -2\)

The Takeaway: The remainder is \(-2\). We didn't have to do any long division!


2. The Factor Theorem

The Factor Theorem is just a special case of the Remainder Theorem. Think about it: if you divide a number and the remainder is zero, it means the number divided perfectly. In algebra, this means the divisor is a factor.

The Rule: If \(f(a) = 0\), then \((x - a)\) is a factor of \(f(x)\).

This is incredibly useful for factorising cubic expressions. If you find a value of \(a\) that makes the whole expression equal zero, you have found a piece of the puzzle!

Step-by-Step Example:

Show that \((x + 1)\) is a factor of \(f(x) = x^3 + 3x^2 - 4\).

Step 1: Identify the value to substitute. For \((x + 1)\), the value is \(x = -1\).
Step 2: Plug it into the function.
\(f(-1) = (-1)^3 + 3(-1)^2 - 4\)
Step 3: Simplify.
\(f(-1) = -1 + 3(1) - 4\)
\(f(-1) = -1 + 3 - 4 = -2\) (Wait, this isn't zero!)
Wait! Let's re-check the syllabus example \(x^3 + 3x^2 - 4\). Let's try \(x = 1\):
\(f(1) = (1)^3 + 3(1)^2 - 4 = 1 + 3 - 4 = 0\).
Conclusion: Since \(f(1) = 0\), then \((x - 1)\) is the factor.

Don't worry if this seems tricky at first! Always double-check your signs. Remember: if the bracket is \((x - a)\), you plug in \(+a\). If the bracket is \((x + a)\), you plug in \(-a\).


3. Important Terminology

To succeed in your exam, you must be comfortable with these three terms:

  • Polynomial: An expression like \(ax^3 + bx^2 + cx + d\).
  • Quotient: The result you get after dividing (the "main" part of the answer).
  • Remainder: What is left over when the division isn't perfect.

Quick Review Box:
If you divide \(f(x)\) by \((ax + b)\), you test the value \(x = -\frac{b}{a}\).
If \(f(-\frac{b}{a}) = 0 \implies (ax + b)\) is a factor.
If \(f(-\frac{b}{a}) = K \implies K\) is the remainder.


4. Common Mistakes to Avoid

Even the best students can make these small errors. Watch out for:

1. Sign Errors: This is the most common mistake. If the divisor is \((x - 3)\), you must substitute \(+3\). If it is \((x + 5)\), you must substitute \(-5\).
2. Brackets with Powers: When substituting a negative number, always use brackets on your calculator. For example, \((-2)^2\) is \(4\), but if you type \(-2^2\) without brackets, some calculators will tell you \(-4\).
3. Missing Terms: If a cubic is written as \(x^3 - 4\), remember that the \(x^2\) and \(x\) terms have coefficients of zero. While this matters more for long division, it's good practice to stay aware of it when substituting.


5. Using the Factor Theorem to Factorise Cubics

The syllabus requires you to factorise expressions like \(6x^3 + 11x^2 - x - 6\) when a factor is provided. Here is how you do it:

The Process:
1. Use the Factor Theorem (substitute the value) to confirm the given factor is correct.
2. Perform algebraic division to divide the cubic by that factor. This will result in a quadratic quotient.
3. Factorise the resulting quadratic using the methods you learned in the "Quadratic function" section.
4. Write down all three factors together.

Did you know? A cubic expression (where the highest power is \(x^3\)) will usually have three linear factors, such as \((x-a)(x-b)(x-c)\).


Summary Key Takeaways

1. Remainder Theorem: To find the remainder when \(f(x)\) is divided by \((x - a)\), just calculate \(f(a)\).
2. Factor Theorem: If \(f(a) = 0\), then \((x - a)\) is a factor of the expression.
3. Division: Once you have a factor, use algebraic division to find the quotient and fully factorise the expression.
4. Accuracy: Always be careful with negative signs during substitution!