Step 1: Indices, Surds and Polynomials
- An index counts how many copies of the base are multiplied together, so \( a^m \times a^n = a^{m+n} \), \( a^m \div a^n = a^{m-n} \) and \( (a^m)^n = a^{mn} \).
- Zero, negative and fractional indices follow from those laws: \( a^0 = 1 \), \( a^{-n} = 1/a^n \) and \( a^{m/n} = (\sqrt[n]{a})^m \).
- A surd is a root left in exact form. Simplify with \( \sqrt{ab} = \sqrt{a}\sqrt{b} \) and clear a surd from a denominator with the conjugate.
- Exam point: \( (x-a) \) is a factor of a polynomial \( f(x) \) exactly when \( f(a) = 0 \), and otherwise \( f(a) \) is the remainder. Watch the sign: dividing by \( (x+2) \) means substituting \( x = -2 \).
Indices: a shorthand for repeated multiplication
Every calculus mark you earn is collected through algebra. You can differentiate perfectly and still lose the mark by mishandling a negative power at the end. So the first step of this guide is the toolkit, starting with indices.
Writing \( 3 \times 3 \times 3 \times 3 \) is tedious, so we write \( 3^4 \). The 3 is the base and the 4 is the index (or power). The index is a count: it tells you how many copies of the base are multiplied together. Hold on to that idea and the laws below are just common sense.
| Law | Example | Why it works |
|---|---|---|
| \( a^m \times a^n = a^{m+n} \) | \( x^2 \times x^5 = x^7 \) | 2 copies and 5 more copies make 7 copies |
| \( a^m \div a^n = a^{m-n} \) | \( x^6 \div x^2 = x^4 \) | dividing cancels 2 of the 6 copies |
| \( (a^m)^n = a^{mn} \) | \( (x^3)^4 = x^{12} \) | 4 groups of 3 copies |
| \( (ab)^n = a^n b^n \) | \( (2x)^3 = 8x^3 \) | every factor in the bracket is copied 3 times |
Why should \( a^0 = 1 \) and \( a^{-1} = \frac{1}{a} \)? Nobody chose these; the pattern forces them. Read down the list \( 3^3 = 27,\ 3^2 = 9,\ 3^1 = 3 \). Each step down divides by 3. Keep going and \( 3^0 = 3 \div 3 = 1 \), then \( 3^{-1} = \frac{1}{3} \), then \( 3^{-2} = \frac{1}{9} \). A negative index does not make anything negative. It means “one over”.
Simplify \( \dfrac{(3a^2b^{-1})^2}{6a^3b^{-4}} \), giving your answer without negative indices.
The trickiest line is \( -2-(-4) = 2 \). Write the subtraction out in full rather than doing it in your head.
Fractional indices and unknown powers
What could \( 16^{1/2} \) mean? Use the power law: \( (16^{1/2})^2 = 16^1 = 16 \). So \( 16^{1/2} \) is the positive number whose square is 16, which is 4. In the same way \( a^{1/3} \) is the cube root of \( a \), and \( a^{m/n} \) is the \( n \)th root raised to the power \( m \).
Take the root first. It keeps the numbers small: \( 32^{3/5} = (\sqrt[5]{32})^3 = 2^3 = 8 \), with no need to work out \( 32^3 \).
(a) Evaluate \( 81^{-3/4} \) and \( \left(\tfrac{16}{81}\right)^{-3/4} \). (b) Solve \( 4^{x+1} = 8^{x} \).
When the bases cannot be matched, as in \( 3^x = 20 \), you need logarithms, which come in Step 5.
An economics example: compound growth. Money earning 4% a year is multiplied by 1.04 each year. So \$5,000 left for 10 years becomes \( 5000 \times 1.04^{10} \approx \$7{,}401.22 \). Simple interest, which adds \$200 a year, would give only \$7,000. The extra \$401.22 is interest earned on earlier interest, and the gap widens every year (Figure 1).
Negative indices run the clock backwards. What is \$10,000 paid in 5 years’ time worth today, if money earns 4%? Its present value is \( 10000 \times 1.04^{-5} \approx \$8{,}219.27 \), because that amount invested now would grow to exactly \$10,000. Bond prices, loan repayments and investment appraisal all use this idea.
Surds and exact answers
Some roots are whole numbers, like \( \sqrt{64} = 8 \). Most are not. \( \sqrt{2} = 1.41421\ldots \) never ends and never repeats, so any decimal is only an approximation. When a question asks for an exact answer, leave the root as a root.
One rule does almost all the work: roots split over multiplication and division.
Both hold for \( a, b \ge 0 \), with \( b \neq 0 \) in the second.
To simplify \( \sqrt{98} \), find the largest square number that divides 98. That is 49, so \( \sqrt{98} = \sqrt{49}\sqrt{2} = 7\sqrt{2} \). Once surds are simplified, collect like surds just as you collect like terms: \( 7\sqrt{2}-4\sqrt{2} = 3\sqrt{2} \), in the same way that \( 7x-4x = 3x \).
By convention, a final answer should not have a surd in the denominator. Removing it is called rationalising the denominator. For a single surd, multiply top and bottom by that surd. For a sum or difference, multiply by the conjugate, which is the same two terms with the middle sign flipped. The difference of two squares, \( (p-q)(p+q) = p^2-q^2 \), then squares the surds away.
(a) Simplify \( \sqrt{98}-\sqrt{32} \). (b) Write \( \dfrac{6}{\sqrt{5}-\sqrt{2}} \) with a rational denominator.
A quick calculator check: \( 6 \div (2.2361-1.4142) \approx 7.30 \) and \( 2(2.2361 + 1.4142) \approx 7.30 \).
Factorising polynomials
A polynomial is a sum of terms of the form \( ax^n \) where each \( n \) is a whole number, such as \( x^3-2x^2-5x + 6 \). Its degree is the highest power, here 3. Expanding turns a product into a polynomial; factorising turns it back into a product. Calculus needs factorising constantly, because a product is zero only when one of its factors is zero. That is how you find where a graph meets the x-axis and, later, where a derivative is zero.
Four patterns cover most of the factorising you will meet:
| Type | Method | Example |
|---|---|---|
| Common factor | Take it out first, always | \( 3x^3-12x = 3x(x^2-4) \) |
| \( x^2 + bx + c \) | Two numbers with product \( c \) and sum \( b \) | \( x^2-x-6 = (x-3)(x + 2) \) |
| \( ax^2 + bx + c \) | Split \( bx \) using two numbers with product \( ac \) and sum \( b \) | \( 2x^2 + 5x-3 = (2x-1)(x + 3) \) |
| Difference of two squares | \( p^2-q^2 = (p-q)(p + q) \) | \( x^2-4 = (x-2)(x + 2) \) |
For the third row, \( ac = 2 \times (-3) = -6 \), and the pair with product \( -6 \) and sum 5 is 6 and \( -1 \). So \( 2x^2 + 6x-x-3 = 2x(x + 3)-(x + 3) = (2x-1)(x + 3) \). “Factorise fully” means keep going until nothing more factorises: the first row finishes as \( 3x(x-2)(x + 2) \).
A cubic or quartic has no “two numbers” trick. You need a way to find one factor, after which the rest is a quadratic. That is the job of the factor theorem.
The factor and remainder theorems
Why is it true? Dividing \( f(x) \) by \( (x-a) \) always gives a quotient \( q(x) \) and a constant remainder \( R \), so \( f(x) = (x-a)\,q(x) + R \). Put \( x = a \): the first part becomes zero, leaving \( f(a) = R \). So the remainder is \( f(a) \), which is the remainder theorem, and \( (x-a) \) divides exactly when that remainder is zero.
Show that \( (x-1) \) is a factor of \( f(x) = x^3-2x^2-5x + 6 \), and hence solve \( f(x) = 0 \).
Check by expanding the comparison: the \( x \) terms give \( -6x + x = -5x \), as required.
\( (x + 2) \) is a factor of \( g(x) = 2x^3 + kx^2-7x-6 \). Find \( k \), then find the remainder when \( g(x) \) is divided by \( (x-1) \).
No division was needed at any point. Two pieces of information (two factors, or a factor and a remainder) give two equations when there are two unknown coefficients.
Practise: 6 questions
Try each one on paper first, then check the worked answer.
EasyQ1. Simplify \( \dfrac{x^5 \times x^{-2}}{x^{1/2}} \). Show worked answer
Divide: \( x^{3-1/2} = x^{5/2} \).
Answer: \( x^{5/2} \), which can also be written \( x^2\sqrt{x} \).
EasyQ2. Evaluate \( 125^{-2/3} \) as a fraction. Show worked answer
Root first: \( \sqrt[3]{125} = 5 \), then \( 5^2 = 25 \).
Answer: \( \dfrac{1}{25} \).
MediumQ3. Simplify \( \sqrt{200}-\sqrt{50} + \sqrt{8} \). Show worked answer
\( \sqrt{50} = 5\sqrt{2} \) and \( \sqrt{8} = 2\sqrt{2} \).
Collect: \( 10\sqrt{2}-5\sqrt{2} + 2\sqrt{2} = 7\sqrt{2} \).
MediumQ4. Express \( \dfrac{3 + \sqrt{2}}{3-\sqrt{2}} \) in the form \( \dfrac{a + b\sqrt{2}}{c} \), where \( a \), \( b \) and \( c \) are integers. Show worked answer
Denominator: \( 9-2 = 7 \).
Numerator: \( (3 + \sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2} \).
Answer: \( \dfrac{11 + 6\sqrt{2}}{7} \).
MediumQ5. Solve \( 9^{x} = 27^{x-1} \). Show worked answer
Equate indices: \( 2x = 3x-3 \), so \( x = 3 \).
Check: \( 9^3 = 729 \) and \( 27^2 = 729 \).
ExamQ6. The polynomial \( p(x) = 2x^3 + ax^2-13x + b \), where \( a \) and \( b \) are constants. It is given that \( (x-2) \) is a factor of \( p(x) \), and that the remainder when \( p(x) \) is divided by \( (x + 1) \) is 18. (a) Find the values of \( a \) and \( b \). [4] (b) Hence factorise \( p(x) \) completely. [3] (c) State the roots of \( p(x) = 0 \). [1] Show worked answer
\( p(-1) = 18 \): \( -2 + a + 13 + b = 18 \), so \( a + b = 7 \). [M1]
Subtracting: \( 3a = 3 \), so \( a = 1 \) and \( b = 6 \). [A1]
(b) \( p(x) = 2x^3 + x^2-13x + 6 = (x-2)(2x^2 + qx-3) \). Comparing \( x^2 \) terms: \( q-4 = 1 \), so \( q = 5 \). [M1 A1]
\( 2x^2 + 5x-3 = (2x-1)(x + 3) \), so \( p(x) = (x-2)(2x-1)(x + 3) \). [A1]
(c) \( x = 2 \), \( x = \tfrac{1}{2} \), \( x = -3 \). [B1]
Total: 8 marks.
Check yourself
1. What is \( 2^{-3} \)?
2. Which of these equals \( \sqrt{50} \)?
3. What is the remainder when \( x^3 + 4x-1 \) is divided by \( (x + 1) \)?
Your score: 0 out of 3
The short version
- An index counts copies of the base. Add indices to multiply, subtract to divide, multiply to raise a power to a power.
- \( a^0 = 1 \), \( a^{-n} = 1/a^n \) and \( a^{m/n} = (\sqrt[n]{a})^m \). Take the root first.
- Simplify surds with \( \sqrt{ab} = \sqrt{a}\sqrt{b} \), and rationalise a denominator such as \( \sqrt{5}-\sqrt{2} \) by multiplying by its conjugate.
- Factorise by taking out common factors, splitting the middle term and spotting differences of two squares. For cubics, find a root \( a \) with \( f(a) = 0 \); then \( (x-a) \) is a factor, and \( f(a) \) is the remainder in general.
- In economics, \( (1 + r)^n \) grows money forward and \( (1 + r)^{-n} \) discounts it back to today.
- The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.
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