12 Algebra Mistakes That Quietly Cost You Marks

These twelve are not a random assortment. Two mathematicians who have kept public catalogues of student errors for decades — Paul Dawkins at Lamar and Eric Schechter at Vanderbilt — arrive at broadly the same short list, and Schechter makes the point that matters: the errors are not independent. Most are a handful of false beliefs, each of which generates a whole family of wrong lines.

So if you correct a single line, you have fixed one question. If you can name the belief, you have fixed every question it was going to spoil, including in topics you have not met yet. A student who understands why $(a+b{)}^2\ne a^2+b^2$ for the right reason will not later write $\ln (a+b)=\ln a+\ln b$, because those are the same error.

An operation $f$ is called additive if $f(x+y)=f(x)+f(y)$ for every $x$ and $y$. A few familiar operations really are — multiplying by a constant, differentiating, taking a limit. Almost nothing else is. Schechter names the belief that everything behaves this way "everything is additive", and treats it as the most common instance of a wider habit he calls "everything is commutative". Four of the twelve below are instances of it.

The misconception is that the exponent distributes over the addition. It does distribute over multiplication — $(ab{)}^2=a^2{b}^2$ is genuinely true — and that neighbouring fact is what makes the error feel plausible. Squaring means multiplying the whole expression by itself, so every term in the first bracket meets every term in the second and both cross terms survive. The same slip appears as $(a-b{)}^2=a^2-b^2$, a collision with the difference of squares: $(a-b)(a+b)=a^2-b^2$ is true, but that is a different product. If you hesitate between the two, expand longhand once; the four-term expansion never lies.

Same belief, inverse operation. A square root is an exponent of $\frac{1}{2}$, so this is mistake 1 read backwards, with the same fix: roots pass through products, not sums. $\sqrt{a^2{b}^2}=ab$ for non-negative $a$ and $b$, and there is no corresponding rule for $a^2+b^2$. The version that costs marks in calculus is simplifying $\sqrt{x^2+1}$ to $x+1$ inside an integral, which turns a hard problem into an easy wrong one.

There is a real rule hiding nearby, and knowing which direction it runs is the whole battle. You may split a fraction across its numerator: $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$ is always valid, since it is just multiplying each term by $1/c$. You may never split one across its denominator: $\frac{c}{a+b}$ is not $\frac{c}{a}+\frac{c}{b}$. Students who learn the first rule and apply it symmetrically produce this error, and it recurs in partial fractions and in simplifying rational expressions.

The log laws are where this misconception does the most damage, because the correct rules look so nearly like the wrong ones. A logarithm turns multiplication into addition, which is precisely why it exists. It does nothing useful to a sum, and $\ln (a+b)$ cannot be broken up at all. The same applies in reverse to exponentials: $e^{a+b}=e^a{e}^b$ is true, and $e^{a+b}=e^a+e^b$ is not. If you remember one sentence about logs, make it this one: they convert between multiplication and addition, so a sum inside a log is a dead end.

Schechter calls sign errors "surely the most common errors of all", and one of the causes he names is not arithmetic at all. It is a writing problem: you think a pair of brackets, do not write them, and then read your own line back a step later as though they were never there.

The minus sign in front of a bracket is a multiplication by $-1$, and it therefore reaches every term inside, not just the first. This is the single most common way to lose an accuracy mark, and it is dangerous because the working after the slip is usually internally consistent — like terms collected correctly, arithmetic clean, nothing for a re-read to catch. Write the distributed form as its own line rather than doing it in your head. One extra line, and the error becomes visible to you and to a marker.

$-3^2$ and $(-3{)}^2$ are different numbers: $-9$ and $9$. The first is the negative of $3^2$, because the exponent binds more tightly than the unary minus. This matters most when substituting a negative value into an expression, and it matters when you type into a calculator, since most of them follow the convention you may be getting wrong. Always write brackets around a negative value you are substituting. It costs one character.

You may cancel a factor common to the whole numerator and the whole denominator. You may never cancel something that is merely one term of a sum. Dawkins lists this as its own category, and Schechter calls it an undistributed cancellation — in effect the reverse of losing invisible parentheses, since you are inserting brackets that are not there.

Compare a cancellation that is legal: $\frac{5x+10}{5}=\frac{5(x+2)}{5}=x+2$. Here $5$ is a factor of the entire numerator, which you can see once it has been pulled out front. That is the practical rule: if you cannot factor it out and see it standing alone, you cannot cancel it. With algebraic fractions the same rule requires you to factor both parts first, and then to record what the factoring removed:

The restriction $x\ne 2$ is not decoration. The original expression is undefined at $x=2$, the simplified one is perfectly happy there, and they are therefore not the same function unless you carry the restriction along. Mark schemes on rational functions frequently allocate a mark for exactly this. The four factoring patterns are what make the first step fast enough to be worth doing every time.

A sign slip costs an accuracy mark; a lost root usually costs a whole answer, and there is nothing on the page for a marker to award follow-through credit to.

Dividing both sides by $x$ silently assumes $x\ne 0$, and if $x=0$ happens to be a solution you have just deleted it. The misconception is treating a variable as though it were a known non-zero number. The fix is mechanical and always available: never divide by an expression containing the unknown. Move everything to one side, factor, and set each factor to zero. That route finds every root by construction, which is why it is the method our equation solver uses rather than cancelling.

Squaring destroys sign information, so undoing it has to restore both possibilities. This is most common in completing-the-square solutions and in geometry problems where a length is found from a squared relationship — and in the geometry case you often do discard the negative, but you discard it because a length cannot be negative, not because it was never there. Those are different acts, and only one of them is mathematics.

Squaring is irreversible: $a=b$ implies $a^2=b^2$, but not the other way round. Any step that is not reversible can introduce solutions that satisfy the transformed equation and not the original one. Schechter reports a colleague who warned students explicitly that an extraneous root would cost half the credit on each problem, and about a third of the class still failed to check.

Note where the second candidate came from. It is not an arithmetic mistake — the factoring is correct and $x=-1$ genuinely solves $x^2-x-2=0$. It is a root of a different equation, one you created when you squared. The same happens when you multiply both sides by an expression containing the unknown, or use a log law that assumes an argument is positive.

Dawkins singles this out as a widespread misconception, and it is worth being precise because mistakes 9 and 11 look contradictory at first. The symbol x​ is a function, and a function returns one value; by convention that value is the non-negative one. When you solve $x^2=16$ you introduce the $\pm$ yourself, because you are asking which numbers square to 16 rather than evaluating a symbol.

The consequence people trip over is that $\sqrt{x^2}=|x|$, not $x$. Check it with $x=-5$: the square is 25, and the principal root of 25 is 5, which is $|-5|$. This is why absolute value signs appear when you simplify radicals with even indices, and why they vanish again when the variable is known to be positive.

Multiplying or dividing an inequality by a negative number reverses the direction, because it reflects the number line about zero. Every other operation you use on equations carries over unchanged, which is what makes this one easy to forget.

There is a nastier version, and it is where good students lose marks rather than careless ones. Consider $\frac{1}{x}>2$. You cannot multiply both sides by $x$, because you do not know its sign, and the direction of the inequality depends on that sign. Split into cases. If $x>0$, multiplying preserves the direction and gives $1>2x$, so $0<x<\frac{1}{2}$. If $x<0$ then $1/x$ is negative and cannot exceed 2, so that case contributes nothing. The solution is $0<x<\frac{1}{2}$ — and the naive one-line approach would have produced $x<\frac{1}{2}$, which wrongly includes every negative number.

Mistake 5 survives proofreading because everything downstream of it is correct: clean arithmetic, properly collected terms, a tidy integer answer. Locate the first bad line here before reading on.

Now run the list against your own hand. Every answer below is one or two lines away; the point is which one you get wrong.

Schechter's warning about checking is worth taking seriously. Read back through the steps you have just written and you are quite likely to make the same mistake again — especially with conceptual errors, because the belief that produced the line is still in place when you read it. Check by a method different from the one that produced the answer.

1. Substitute a number. Put $x=2$ into the original expression and into your simplified one and compare. Avoid $x=0$ and $x=1$, which hide too many errors. This one check catches mistakes 1, 2, 3, 4, 5, 7 and 12. It cannot catch mistake 6, because that error happens during substitution itself — which is why the brackets go in every time.
2. Count your roots. A quadratic has two, a cubic has three, counting repeats. If you finished a quadratic with one root, either it is a genuine repeated root — and you should be able to say so — or you divided by a variable or dropped a plus-or-minus. That covers mistakes 8 and 9.
3. Write the domain first, check it last. Before you start, note every value that would make a denominator zero or put a negative under an even root. At the end, cross off any answer on that list, and check every answer if any step squared both sides or multiplied by an expression containing the unknown. That is mistake 10.
4. Re-read only the brackets. Not the whole solution. Scan for lines where a minus sign meets an opening bracket and verify each one in isolation. Narrowing the target is what makes this different from reading it all again.

None of this helps if you never find out you were wrong, which is the argument for checking against a full worked solution rather than an answer key. An answer key tells you that something went wrong; a complete set of steps tells you which line. Compare yours line by line and stop at the first divergence — that line is the entry for your error log, in the format described in our guide to studying for a maths test. If the divergence is in the last few lines rather than the algebra, the likely culprit is presentation, which we cover in why the right answer still loses marks.