Understanding Algebraic Notation

In algebra, when a number is written directly next to a letter with no sign between them, it always means multiplication. This is the “invisible multiplication sign” convention — we drop the × sign because it looks too similar to the letter \( x \), which would cause confusion. To verify, substitute a value: if \( a = 4 \), then \( 3a = 3 \times 4 = 12 \). The “3” is called the coefficient — it tells you how many lots of \( a \) you have.

A further way to convince someone: \( 3a = a + a + a \). If each \( a \) is worth 4, then \( a + a + a = 4 + 4 + 4 = 12 \), which matches \( 3 \times 4 = 12 \). The invisible multiplication is a shorthand for repeated addition.

Adding \( a \) three times gives 3 lots of \( a \), which we write as \( 3a \). The “3” in \( 3a \) is the coefficient — it tells you how many \( a \)’s are being added together. But \( a^3 \) means \( a \times a \times a \) — the “3” in \( a^3 \) is the exponent, telling you how many times \( a \) is multiplied by itself. This is the “confusing coefficients with exponents” misconception — the position of the number completely changes its meaning.

Substitution proves it. If \( a = 2 \): \( a + a + a = 2 + 2 + 2 = 6 \), and \( 3a = 3 \times 2 = 6 \) ✓. But \( a^3 = 2 \times 2 \times 2 = 8 \neq 6 \). If \( a = 5 \): \( 3a = 15 \), but \( a^3 = 125 \). The exponent version grows much faster because multiplication compounds — this is why the distinction matters.

\( 4a \) means “4 lots of \( a \)” and \( 3b \) means “3 lots of \( b \)”. Since \( a \) and \( b \) represent different unknown values, these are unlike terms and cannot be combined. This addresses the “combining unlike terms” misconception — the belief that \( 4a + 3b = 5ab \) or \( 7ab \). You can only collect terms that have exactly the same variable part. For example, \( 4a + 3a = 7a \) works because both are “lots of \( a \)”, but \( 4a \) and \( 3b \) are measuring different things.

Testing with values confirms this: if \( a = 5 \) and \( b = 2 \), then \( 4a + 3b = 20 + 6 = 26 \). If it simplified to \( 7ab \), we’d get \( 7 \times 5 \times 2 = 70 \neq 26 \). The expression \( 4a + 3b \) is already as simple as it can get.

Since \( 1 \times \) anything = that thing, multiplying \( x \) by 1 doesn’t change its value. So \( 1x = 1 \times x = x \). This is the “invisible coefficient” misconception — students may think \( x \) has “no coefficient” rather than a coefficient of 1. In algebra, we follow the convention of dropping the coefficient when it’s 1, because writing it adds no information. This applies to any term: \( 1x^2 \) is written as \( x^2 \), and \( 1ab \) is written as \( ab \).

The test: when simplifying \( 5x \; – \; 4x \), you need to recognise that \( x \) has a coefficient of 1 to get the answer. We calculate \( 5x \; – \; 4x = (5 \; – \; 4)x = 1x = x \). If you didn’t know \( x \) meant \( 1x \), you might not see where the answer comes from.

You cannot simply “remove” the letter \( a \) from the term \( 5a \) by subtracting \( a \). We must remember that \( a \) is shorthand for \( 1a \). Therefore, the calculation is \( 5a – 1a = 4a \). Think of it in terms of objects: if you have 5 apples and take away 1 apple, you have 4 apples left, not the number 5.

If the answer were 5, then no matter what \( a \) is, the result would be 5. Let’s test this: if \( a = 10 \), then \( 5a – a = 50 – 10 = 40 \). This is definitely not 5!

Example: \( 5x + 3 \)

Another: \( 5x \; – \; 7 \)

Creative: \( 5x + 2y \; – \; 1 \) — the coefficient of \( x \) is still 5 even in a multi-variable expression with other terms present. The 2 is the coefficient of \( y \), and −1 is a constant.

Trap: \( x + 5 \) — the 5 here is a constant term (a number on its own), not the coefficient of \( x \). This is the “confusing coefficient with constant” misconception. The coefficient of \( x \) in this expression is actually 1, because \( x \) means \( 1x \). Students often confuse “the number near \( x \)” with “the number multiplying \( x \).”

Example: \( 2n + 5 \)

Another: \( 2x + 5 \)

Creative: \( 5 + 2n \) — this is the same expression rearranged, since addition is commutative. Writing the constant first is perfectly valid and means exactly the same thing.

Trap: \( 2(n + 5) \) — this looks similar, but the brackets change the meaning entirely. This expression means “add 5 first, then multiply the total by 2.” Testing with \( n = 3 \): the correct version gives \( 2(3) + 5 = 11 \), but the trap gives \( 2(3 + 5) = 2 \times 8 = 16 \). The “brackets change order of operations” misconception — students confuse \( 2n + 5 \) with \( 2(n + 5) \).

Example: \( 2x + 2x \)

Another: \( x + x + x + x \)

Creative: \( 10x \; – \; 6x \) — using subtraction to arrive at \( 4x \). Or \( 4 \times x \), since \( 4x \) literally means “4 multiplied by \( x \).”

Trap: \( x^4 \) — students sometimes confuse the coefficient (the multiplier in front) with the exponent (the small number above). But \( x^4 \) means \( x \times x \times x \times x \), while \( 4x \) means \( x + x + x + x \). Testing with \( x = 2 \): \( 4x = 8 \), but \( x^4 = 16 \). With \( x = 3 \): \( 4x = 12 \), but \( x^4 = 81 \). This is the “coefficient vs exponent” misconception.

Example: \( a + 3 \) and 5. When \( a = 2 \): \( 2 + 3 = 5 \) ✓. But when \( a = 1 \): \( 1 + 3 = 4 \neq 5 \) ✗.

Another: \( a^2 \) and \( 2a \). When \( a = 2 \): \( 2^2 = 4 \) and \( 2 \times 2 = 4 \) ✓. But when \( a = 3 \): \( 3^2 = 9 \) and \( 2 \times 3 = 6 \) ✗.

Creative: \( 3a \) and \( a + 4 \). When \( a = 2 \): \( 3(2) = 6 \) and \( 2 + 4 = 6 \) ✓. But when \( a = 1 \): \( 3(1) = 3 \) and \( 1 + 4 = 5 \) ✗. Or try \( a^2 \; – \; 2 \) and \( a \). When \( a = 2 \): \( 4 \; – \; 2 = 2 \) ✓. When \( a = 3 \): \( 9 \; – \; 2 = 7 \neq 3 \) ✗.

Trap: \( 2a \) and \( a + a \) — these ARE equal for every value of \( a \) (since adding \( a \) twice always gives \( 2a \)), so they don’t satisfy the condition. Students may think “they look different so they must not always be equal,” but these are genuinely equivalent expressions. This targets the “equivalence vs coincidence” misconception.

Multiplication is commutative, which means \( a \times b = b \times a \) for all values of \( a \) and \( b \). Since \( ab \) means \( a \times b \) and \( ba \) means \( b \times a \), these are always equal. In algebra, we conventionally write letters in alphabetical order (\( ab \) rather than \( ba \)), but this is just a convention for tidiness — the expressions mean exactly the same thing.

Students may fall for the “order always matters” misconception — confusing multiplication (where order doesn’t matter) with subtraction or division (where it does). The key insight is that multiplication and addition are commutative, but subtraction and division are not.

This depends on the value of \( a \). TRUE case: when \( a = 3 \), \( a^2 = 9 > 3 \) ✓. More generally, \( a^2 > a \) whenever \( a > 1 \) or \( a < 0 \) (for example, when \( a = -3 \), \( a^2 = 9 > -3 \)). FALSE case: when \( a = \frac{1}{2} \), \( a^2 = \frac{1}{4} < \frac{1}{2} \) ✗. For any value between 0 and 1, squaring makes the number smaller. And when \( a = 1 \), \( a^2 = 1 = a \) (equal, not greater).

The misconception here is “squaring always makes things bigger.” This is true for whole numbers greater than 1, which is what students encounter most often, so the pattern feels universal. But squaring fractions makes them smaller — a crucial insight for understanding powers.

Setting them equal: \( 5 \; – \; a = a \; – \; 5 \) gives \( 10 = 2a \), so \( a = 5 \). TRUE case: when \( a = 5 \), both expressions give 0 ✓. FALSE case: when \( a = 3 \), \( 5 \; – \; 3 = 2 \) but \( 3 \; – \; 5 = -2 \) ✗. When \( a = 8 \), \( 5 \; – \; 8 = -3 \) but \( 8 \; – \; 5 = 3 \) ✗.

This targets the “subtraction is commutative” misconception — the belief that the order doesn’t matter when subtracting. Students who think \( a \; – \; b = b \; – \; a \) for all values often make errors in simplifying expressions. In fact, swapping the order reverses the sign: \( 5 \; – \; a \) and \( a \; – \; 5 \) always produce results with opposite signs (except when both are zero).

For any real value of \( a \), multiplying \( a \) by itself always gives a result that is zero or positive. If \( a \) is positive, then positive × positive = positive. If \( a \) is negative (say \( a = -3 \)), then \( a^2 = (-3) \times (-3) = 9 \) (positive, because negative × negative = positive). If \( a = 0 \), then \( a^2 = 0 \) (which is not negative).

The key misconception here is confusing \( a^2 \) with \( -a^2 \) or believing \( (-3)^2 = -9 \). Students often think the negative sign “stays” after squaring. The distinction is: \( (-3)^2 \) means \( (-3) \times (-3) = 9 \), while \( -3^2 \) means \( -(3 \times 3) = -9 \). The brackets make all the difference. But the expression \( a^2 \) with no negative sign in front can never be negative (assuming real numbers).

In algebra, the fraction bar is used to represent division. This is a matter of definition rather than calculation. Writing \( \frac{a}{b} \) is the standard way to express \( a \) divided by \( b \). Often, students view fractions and division sums as two separate concepts, but they are mathematically identical.

The student has added the coefficients instead of multiplying them (the “default to addition” misconception). When you multiply algebraic terms, you multiply the number parts together: \( 2 \times 3 = 6 \). Then the variables are written together: \( a \times b = ab \). So the correct answer is \( 6ab \).

We can see this using an area model. If you have a rectangle with height \( 2a \) and width \( 3b \), you can split it into a grid of 6 smaller rectangles, each with area \( ab \):

The answer is correct, but the reasoning contains the “letters as labels” (or “fruit salad algebra”) misconception. The student treats \( p \) as an abbreviation for “pen” (an object) rather than understanding that \( p \) represents a number — specifically, the price of one pen in pence. Similarly, \( r \) is not “a ruler” but “the price of one ruler in pence.”

This reasoning happens to produce the right answer here, but it breaks down with more complex relationships. For example, if “each ruler costs 10p more than each pen,” the correct equation is \( r = p + 10 \). Under the student’s logic, “ruler = pen + 10” makes no sense — you can’t add 10 to an object. Letters in algebra always represent numbers, not things. The expression \( 3p \) means “3 times the price of one pen,” not “3 pens.”

The student has confused exponents with multiplication (the “power means multiply” misconception). The expression \( a^2 \) means \( a \times a \) (\( a \) multiplied by itself), not \( a \times 2 \). The small raised “2” is an exponent — it tells you how many times the base appears in the multiplication. So when \( a = 5 \): \( a^2 = 5 \times 5 = 25 \), not \( 5 \times 2 = 10 \).

A helpful contrast: \( 2a \) and \( a^2 \) look similar but mean very different things. When \( a = 5 \): \( 2a = 2 \times 5 = 10 \), while \( a^2 = 5 \times 5 = 25 \). The student has actually calculated \( 2a \), not \( a^2 \). When \( a = 3 \): \( 2a = 6 \), but \( a^2 = 9 \). The difference grows rapidly — when \( a = 10 \): \( 2a = 20 \), but \( a^2 = 100 \).

The student has treated all numbers as interchangeable, ignoring whether they are coefficients or constants (the “all numbers are alike” misconception). The expression contains two types of terms: variable terms (\( 5a \) and \( -2a \), both involving \( a \)) and a constant term (3, which is just a number on its own). You can only combine like terms: \( 5a \; – \; 2a = 3a \), and the 3 stays separate as a constant. The correct answer is \( 3a + 3 \).

Testing: if \( a = 4 \), the student gets \( 6a = 24 \). But \( 5(4) + 3 \; – \; 2(4) = 20 + 3 \; – \; 8 = 15 \). The correct answer \( 3a + 3 \) gives \( 3(4) + 3 = 12 + 3 = 15 \) ✓. The constant “3” is not “3 lots of \( a \)” — it is the standalone number 3. Combining it with the coefficients of \( a \) treats it as something it isn’t.

The student has failed to use brackets when substituting a negative number. When we substitute \( x = -4 \), we must square the entire value of \( x \). This should be written as \( (-4)^2 \), which means \( (-4) \times (-4) = 16 \).

The student effectively calculated \( -4^2 \), which means “the negative of \( 4^2 \)” (or \( -1 \times 4^2 \)). This is a subtle but vital difference in notation. The negative sign is attached to the number 4, so it gets squared too.