Math 115
Math 105
The top fourteen algebra errors made by struggling students:
This document gives a detailed description of the most common algebra errors made by
struggling students. With each example the faulty reasoning behind the error is explained and
then a correct example will follow. Read through this document and see if you recognize any of
these errors. Quizzes or exams may be returned to you with error statements such as "E3" or
"E7" written upon it. This document will explain those errors in detail. By identifying these
errors early in the semester you will help prevent their occurrence on quizzes and exams. Identify
and negate!
Introductory Note: Most of these algebraic errors are due to confusing the
operations of Factors and Terms. If you can clearly see a mathematical expression as an
arrangement of factors and terms, then their respective operations should remain true and
discourage these errors from occurring.
FACTORS: Values that are connected by multiplication or division.
TERMS:
Values that are connected by addition or subtraction.
Factors within terms:
In the first term we can 'see' either 2 factors: 4  x 2
or 3 factors: 4  x  x
2
2
or 4 factors: 2  2  x  x
4 x  15xy  22 y
3 terms
Terms within factors:
The first factor has three terms: 6x, -2y, and 5.
6x  2 y  53  5x7 x  y 
And yes, the first term of 6x is made
up of 2 factors: 6  x .
3 factors
Turn the page to examine the most common algebra errors
Math 115
Math 105
E1.)
The top fourteen algebra errors made by struggling students:
Illegal canceling: Canceling the factors from only some of the terms of a fraction.
If you are going to cancel, a cancelled factor must come from each and every term of the
fraction. For Example:
ERROR:
3x  5 3x  5 x  5


6
6
2
fraction’s term of ‘-5’, then Hands Off! The fraction
is fully reduced and nothing can be cancelled.
Terms
CORRECT:
14 x  12w 2  7 x  2  6w 2  7 x  2  6w 7 x  6w



10  6 y
2  5  2  3y
2  5  2  3y
5  3y
The four terms of the fraction each have a factor of ‘2’ which can
be cleanly cancelled to reduce the fraction. And yes, it would be a
mistake to try to cancel the ‘3’ with the ‘6’ of the reduced fraction.
Terms
E2.)
If a factor of 3 can not be evenly divided into the
Partial distribution of a factor: Neglecting to distribute a factor to every term of a
quantity. When a factor is positioned in front of a quantity it must be distributed to each
and every term. For Example:
ERROR:
 2x  4 y  7  2x  4 y  7
ERROR:
 33x  5 y  8  9x  15 y  24
CORRECT:
 47 x  6 y  5  28x  24 y  20
E3.)
The negative sign was distributed to the first
term only.
The factor of ‘3’ was distributed to every
term but the negative sign again was only
distributed to the first term.
Distribution of exponents over terms: An exponent is a tally of the number of times the
number ‘1’ has been multiplied by a particular base. If the exponent is negative it is a
tally of the number of times the number ‘1’ has been divided by a base. But since the
exponent is a tally is of the operations of multiplication and division it can not be
distributed over terms that are joined by addition and subtraction. For Example:
ERROR:
x  42  x 2  16
CORRECT:
 x  42   x  4   x  4  x 2  8 x  16
If ‘x + 4’ is the base then it must be multiplied by itself
twice as indicated by the exponent of ‘2’.
The 2 factors must be fully distributed, or ’FOIL’ed
E3.) Continued…
Math 115
Math 105
The top fourteen algebra errors made by struggling students:
E3.) Continued; Distributing exponents over terms.
This type of error frequently occurs when solving equations containing radical
expressions. Solve the following equation.
CONSIDER: 1  2 x  7  x  3
12 
ERROR:

2x  7
 
2
x3

2
In an attempt to clear the radicals in one smooth step, our ‘struggling solver’ has
squared each term of the equation. This approach is intuitive, direct and
dreadfully wrong. Here’s why: Exponents are a tally of the number of times a
particular base has been multiplied by itself. We cannot distribute this tally over
addition and subtraction. The equation will not remain true.
SAME ERROR, A
REVEALING EXAMPLE:
5 2  25
(2  3) 2  25
A true equation.
Still true, yes?
2 2  3 2  25
4  9  25
13  25
ERROR! Same glitch as in the
example above. Exponents were
distributed over terms.
CORRECT APPROACH: (For solving the radical equation above.)
1  2x  7  x  3
Solve for x.
2x  7  1  x  3

2x  7
  1 
x3


2
First isolate the busiest radical.

2
Square both sides of the equation.
2x  7  1  x  3  1  x  3

Distribute terms on the right side of equation.
2x  7  1  2 x  3  x  3
2 x  3  2x  7  x  1  3
Again, isolate the radical.
2 x3  x3
Collect like terms.
2
x3

2
  x  3
2
4( x  3)  x 2  6 x  9
x 2  2x  3  0
Square both sides of the equation.
Distribute and set equation equal to zero.
Solution: x  3 , x  1
Math 115
Math 105
E4.)
The top fourteen algebra errors made by struggling students:
Distributing roots over terms: A root, as in a square root or cube root, is also an
1
1
exponent: 4  4 2 , 3 27  27 3 . Because the root as an exponent is a tally of the operations of
multiplication and division the root can not be distributed over terms that are joined with
addition and subtraction. See E3. For Example:
ERROR:
x 2  49  x 2  49  x  7
CONSIDER:
ERROR:
25  5
Because the terms are joined by
addition the root can not be reduced.
A true statement.
9  16  5
Still true.
9  16  5
A faulty move.
3 4  5
ERROR:
Glitch! 7  5
Note: If the root contains only products and quotients then we can distribute the root.
Because a root like an exponent is a tally of the operations of multiplication and division, then
distribution to products and quotients is a legal move.
CONSIDER:
36  6
A true statement.
49  6
Still true.
4 9 6
Yes, a good move.
23  6
6=6
E5.)
All true because the root was distributed over
factors, not terms.
Letting a term drop to zero when factored from a quantity: Factoring is akin to
division and when we divide a number by itself the result is not zero, rather it is the
number ‘1’.
ERROR:
x  3x
2

 4 xy  x  3x  4 y 
Pulling a common factor of ‘x’ from each term of the quantity the factor of x was
assumed (incorrectly) to drop out as a zero. If the quantity has 3 terms within its
parenthesis there should still be 3 terms after pulling out the common factor. As a check
try distributing the pulled factor to be sure that you can reproduce the original quantity.
CORRECT:
x  3x
2

 4 xy  x  1  3x  4 y 
Math 115
Math 105
E6)
The top fourteen algebra errors made by struggling students:
Collecting unlike terms.
ERROR: 2x + 5  7x
Both terms must hold an identical factor of 'x' before we
can add them together.
terms:
SAME ERROR:
3
1 1
Ct   Ct
4
2 4
This error tends to occur when working with new
variables. We must 'see' C t as just another variable.
CORRECT: 2x + 5 = 2x + 5
CORRECT:
E7.)
3
1 3
1
Ct   Ct 
4
2 4
2
CORRECT:
2x + 5x = 7x
CORRECT
3
1
1
Ct  Ct  Ct
4
2
4
Unlike terms remain untouched.
But like terms can be combined.
Dividing an equation by the variable one is solving for. This is a subtle error that is not
wrong in terms of the mechanics used, rather it is in the choice of steps used to solve an
equation. When we divide an equation by the variable we are solving for we unwittingly
‘divide out’ part of the solution. To avoid this error the first step to solve an equation
should always be to set the equation equal to zero.
ERROR EXAMPLE:
Solve for x:
x 2  4x
Divide by x:
x=4
This is a true statement, but it does not include every
solution. x = 0 is also a solution but it was ‘divided out’.
CORRECT APPROACH:
Solve for x:
x 2  4x
Set the equation equal to zero.
x 2  4x  0
Factor.
x( x  4)  0
Solve using zero products.
x = 0, x = 4
We have all the solutions to the equation.
Math 115
Math 105
E8.)
The top fourteen algebra errors made by struggling students:
Solving for variables when simplifying expressions: When simplifying expressions
such as reducing fractions or factoring polynomials the goal is to find a more compact
form for the expression and then we stop! If an expression is not set equal to a specific
value then we can not force a solution by setting the expression equal to zero or one.
For Example:
CONSIDER: Factor:
x 2  2 x  15  x  3x  5
Done! The polynomial has been factored.
FALSE ASSUMPTION:
x  3x  5  0
ERROR:
x-3 = 0
x+5=0
ERROR:
x=3
x = -5
CONSIDER: Reduce:
3x
x

18 x  21 6 x  7
Rarely written down, a jump to the next step
is made from this false assumption.
In a false assumption, the polynomial has
been set to zero, which means its resulting
factors will also be equal to zero. From this
assumption we incorrectly solve for x.
Done! A factor of ‘3’ has been cancelled
from each term of the fraction.
x
1
6x  7
FALSE ASSUMPTION:
Rarely written down, a jump to the next step
is made from this false assumption.
ERROR:
x = 6x – 7
ERROR:
– 5x = – 7
ERROR:
x=
E9.)
7
5
The fraction has been set to the number ‘1’ and in a
jump to the next step the denominator has been
cleared by multiplying both sides of an assumed
equation by the denominator of ‘6x-7’. From this
false assumption we incorrectly solve for x.
Neglecting to distribute exponents to integer factors: When raising a quantity of
several factors to an exponent each and every factor must be raised to that exponent. The
mistake of not distributing the exponent to integer factors may occur because an integer is
often written without an exponent. When we write the number ‘7’ we don’t include an
exponent of the first power but it is there none the less: 7  7 1 . For Example:
ERROR:
7 x 2
 7x2
ERROR:
3  w
3
 x4

 3w 6 x 8
The factor of ‘3’ was not squared.
CORRECT:
5  p
2
 x3

 5 4 p 8 x12
All three factors are raised to the power of ‘4’.
2
4
The factor of ‘7’ must also be squared.
Math 115
Math 105
The top fourteen algebra errors made by struggling students:
E10.) Pulling only the leading coefficient from a polynomial in an attempt to factor: If we
are going to factor a polynomial by pulling out a common factor, then this common factor
must come from each and every term of the polynomial. For Example:


ERROR:
3x 2  12 x  51  3  x 2  12 x  51
CORRECT:
3x 2  12 x  51  3x 2  3  4 x  3  17 The factor of ‘3’ is pulled from each and
= 3  ( x 2  4 x  17)
every term.
The second and third term were not factored.
If we wish to pull the leading coefficient from a polynomial and it will not divide
evenly from each term then we still must show that every term has had a factor of three
pulled from its’ value.
CONSIDER: 3 x 2  7 x  40
CORRECT:
The coefficients of ‘7’ and ‘40’ can not be divided evenly
by a factor of ‘3’. We must form fractions of thirds.
7 x 40 

3x 2  7 x  40  3   x 2 
 
3
3 

E11.) Confusing the mechanics of exponents about a fraction bar: There is short hand rule
that is often used when working with multiple factors within a fraction: “A factor may
cross the fraction bar if we change the sign of the exponent”. This is a fine rule as long as
we are clear about what a factor is and that the change of sign is applied to exponents
only. Most of these errors occur because the integers are not seen as separate factors but
rather as companion coefficients that ‘ride’ with their adjacent variables. Worse, a
negative factor is read as having a negative exponent. For Example:
ERROR:
y
 7 xy
7 x 1
The factor of ‘7’ has its own exponent of ‘1’, it can not be placed
in the numerator unless the sign of its’ exponent is changed.
CORRECT:
y
xy

1
7
7x
ERROR:
y 3
6x
 3
1
 6x
y
Or as suggested above:
y
 7 1 xy
1
7x
The factor of ‘  6 ’ was confused with 6 1 .
A negative factor is not a negative exponent.
The factor of ‘  6 ’ has its own exponent of ‘1’.
CORRECT:
y 3
x
 3
1
 6x
6y
Or alternatively:
y 3
 6 1 xy 3 .
 6 x 1
Math 115
Math 105
The top fourteen algebra errors made by struggling students:
E12.) Forcing the zero product rule on non-zero products.
Given a quadratic equation is set to zero we can solve for it's variables by
factoring and then setting each factor of the polynomial to zero. This is the zero product
rule. If products result in zero we can assume that each product holding a variable can be
set to zero to solve for the quadratic equation. We can do this simply because anything
times zero is zero. We create zero products.
CORRECT
EXAMPLE:
6 x 2  x  12  0
Let's solve for x by using zero products.
(2x – 3)(3x + 4) = 0
Factoring gives us a zero product:
2 quantities multiplied together that equal zero.
(2x – 3) = 0
x
(3x + 4) = 0
3
2
x
We set each factor to zero and solve for x.
4
3
Done!
The zero product rule only works for factored quantities set to zero. No other value will work.
Consider the error of forcing this rule on non-zero products:
CONSIDER:
6 x 2  x  12  3
The same polynomial is set equal to 3.
CAREFUL!
(2x – 3)(3x + 4) = 3
We should have subtracted 3 from both sides to set
the equation equal to zero.
ERROR!
(2x – 3) = 3
(3x + 4) = 3
ERROR!
x=3
x=
Serious Error. We can't solve for non-zero products.
1
3
Not Valid
6 x 2  x  12  3
CORRECT APPROACH:
6 x 2  x  15  0
(2x + 3)(3x - 5) = 0
x
3
2
x
Set the equation to zero.
Factor the equation.
5
3
And solve for x. Done!
Math 115
Math 105
The top fourteen algebra errors made by struggling students:
E13.) Distributing factors over terms through multiple quantities.
ERROR:
23  g 4  z   6  2g 8  2z 
The factor of '2' was distributed incorrectly
through both quantities, instead of just one. The
quantities (the expression contained within a set of
parentheses) must be seen as factors. And we can
not distribute factors over factors, we can only
distribute factors over terms.
3 Factors
CONSIDER: 2(x)(y) = 2xy  2x2y
In the form above, the error is rare only because we all recognize that 3 factors multiplied
together will not have any operations of distribution. If we can just recognize that (3  g ) is just a
big 'x' factor and that ( 4  z ) is just a big 'y' factor this error would not occur.
To distribute that factor of 2 correctly we must choose one quantity to receive it. Either
one is fine as long as it is the only one. Consider the following correct examples.
CORRECT:
23  g 4  z   6  2g 4  z 
ALSO CORRECT:
23  g 4  z   3  g 8  2z 
E14.) Pulling the inputs from functions as if they were common factors.
This is a horrid error that violates the meaning of a function. A function is a rule
of correspondence between 2 variables: Every input will have exactly one output that is
generated according to a specific rule. For example let f ( x)  2 x  3 . The rule of
correspondence for this function says that all inputs, x, will be doubled and then 3 will be
added to arrive at the output, f. The error results from seeing f (x) as two random
variables that are being multiplied together instead of a coupled unit that defines a
function.
FUNDAMENTAL ERROR EXAMPLE:
Given f ( x)  f ( x 2 )  0
Solve for x.
SERIOUS ERRORS: f ( x  x 2 )  0
x[ f  f ( x)]  0
The function is being treated as 2 random variables multiplied together, An awful error.
E14) Continued
Math 115
Math 105
The top fourteen algebra errors made by struggling students:
E14.) Pulling the inputs from functions as if they were common factors. (Continued)
If you think that this is an unlikely mistake consider that these errors occur when
one is working with logarithmic and trigonometric functions. The function's state
variable of ‘ln’, ‘sin’ or ‘cos’ are not readily seen as defining functions the way f and g
are when presented as f(x) and g(x). Yet all are functions and can not be expressed
without their input variables and those inputs can not be pulled as if they were common
factors. On the next page are a few examples of this fundamental error:
Solve for x given:
ln( x)  ln( x 2 )  27
ERROR:
ln( x  x 2 )  27
The rule of logs correctly says: ln( x 3 )  27
EVEN WORSE:
x(ln  ln( x))  27
Again the ‘ln’ is being treated as a variable.
And the natural log cannot exist without an
input variable.
Solve for x given:
sin( x)  cos( x 2 )  1
ERROR:
x[sin  cos( x)]  1
This is just awful. The input variable of x cannot be
pulled as if it were a common factor sitting within
parenthesis. And sine as a function can not exist
without an input variable. The parenthesis that
follow sine and cosine signal that each are functions
and the values contained within those parenthesis
are input variables, not multiplied quantities.
CORRECT APPROACH: Use the rule of correspondence to expand each function.
Given f ( x)  f ( x 2 )  0
Solve for x, where f ( x)  2 x  3
2x + 3 + 2 x 2  3  0
2x 2  2x  6  0
Collect like terms.
x2  x  3  0
Solve by using the quadratic formula or by completing the square.
x
 1 i 11

2
2
Yes!