Finding the Error: We frequently make algebra mistakes due to common confusions, such as expanding and simplifying rules, fractions, indices, and equations, which lead to incorrect answers. Furthermore, these error patterns are fundamental and easily corrected. Check if you make similar mistakes because, surprisingly, even students at the university level make these silly mistakes.

There are many chances of making errors in finding the factorisation of the polynomials, such as expanding the brackets and distributing common factors will lead to wrong calculations. Knowing proper rules such as BODMAS and simplifying brackets will help reduce errors.

Factorisation and Errors

In the algebraic expression \(5xy + 3\), the term \(5xy\) has been formed by the factors \(5,x,\) and \(y\) i.e., \(5xy = 5 \times x \times y\). It is worth noting that the factors \(5,x,\) and \(y\) that are factors of \(5xy\) cannot be expressed as a product of their factors. Hence, we can call \(5,x,\) and \(y\) the ‘prime’ factors of \(5xy\). In algebraic expressions, the word ‘irreducible’ is used instead of ‘prime’. Therefore, \(5xy\) is the irreducible form of \(5xy\).

Thus, writing the expressions as a product of factors that are in irreducible form is called factorisation.

Note that \(5 \times \left( {xy} \right)\) is not an irreducible form of \(5xy\) since the factor \(xy\) can be further expressed as a product of \(x\) and \(y\), i.e., \(xy = x \times y\). Here, writing \(5 \times \left( {xy} \right)\) is an error in factorisation. We can discuss more types of such common errors in this article.

Finding Error: Division by Zero

We know that \(\frac{0}{2} = 0\). The issue is that far too many students also say that \(\frac{0}{2} = 0\) or \(\frac{0}{2} = 2\). Remember that division by zero is undefined. You cannot divide by zero.

This simple error led us to something we knew was not true. But in most cases, the answer will not be obviously incorrect. It is essential to keep in mind that it is not always explicit that you are dividing by zero, as in this example.

Remember, you CAN NOT divide by zero.

Finding Error: Correct Use of Parenthesis

It is probably the most perplexing mistake that causes an error in factorisation. Here are a few common errors that students make.

• The first mistake is that students will forget to include and decide that parenthesis aren’t necessary at certain steps. The problem is that they frequently forget about them in the following steps of the problem.
• Another error is that students do not always understand the significance of parenthesis.

Example 1: Square of \(4x\)
Solution: Most of the students write \({\left( {4x} \right)^2}\) as \(4{x^2}\), which is incorrect.
The correct answer is \({4^2}{x^2} = 16{x^2}\)

• Take note of the significant difference between these two cases. In the incorrect case, we have squared only \(x\) and not the \(4\), which led to the error.
• But, in the correct case, the whole quantities inside the parenthesis are squared. So, we squared both \(x\) and \(4\).

Example 2: Subtract \(4x – 5\) from \({x^2} + 3x – 5\).
Solution: \({x^2} + 3x – 5 – 4x – 5 = {x^2} – x – 10\) which is wrong.
The correct statement is obtained by using the parenthesis
\({x^2} + 3x – 5 – (4x – 5) = {x^2} + 3x – 5 – 4x + 5\)
\({x^2} + 3x – 5 – (4x – 5) = {x^2} – x\)

• Take care to distinguish between the two. In the first case, not use parenthesis around the \(4x – 5\), but used in the second. Since we are subtracting a polynomial, we must ensure that we subtract the whole polynomial. The only way to ensure that we do it correctly is to surround it with parenthesis.
• Again, this is one of those errors where students are aware that the parenthesis should be present but fail to include them, then forget about them and perform the subtraction incorrectly.

Finding Error: Improper Distribution

Use the distributive property with caution. There are two significant errors that we encounter.

Example 1: Multiply \(4\left( {2{x^2} – 10} \right)\).
Solution: \(4\left( {2{x^2} – 10} \right) = 8{x^2} – 10\), which is incorrect.
\(4\left( {2{x^2} – 10} \right) = 8{x^2} – 40\) is the correct one.

• Ensure that the \(4\) is distributed throughout the parenthesis.
• Students frequently multiply the first term by \(4\) and ignore the second term.
• Students will remember to do the distribution correctly.

Example 2: Multiply \(3{\left( {2x – 5} \right)^2}\).
Solution: \(3{(2x – 5)^2} = 36{x^2} – 180x + 225\), which is incorrect.
The correct answer is obtained as follows:
\(3{(2x – 5)^2} = 3\left( {{{(2x)}^2} – 2(2x)(5) + {5^2}} \right)\)
\( = 3\left( {4{x^2} – 20x + 25} \right)\)
\( = 12{x^2} – 60x + 125\)

• Remember to perform exponentiation before distributing any coefficients through the parenthesis.
• Remember that when you multiply the expression enclosed in a bracket by a constant (or variable) outside, you must multiply each term of the expression by the constant (or the variable).

Finding Error: Correct Division

These errors are classified into two types.

1. Solving equations
2. Simplifying rational expressions

First, consider simplifying rational expressions.

Example: Simplify \(\frac{{3{x^3} – x}}{x}\).

Correct Solution	Incorrect Solution
\(\frac{{3{x^3} – x}}{x} = \frac{{x\left( {3{x^2} – 1} \right)}}{x}\) \({ = 3{x^2} – 1}\)	\(\frac{{3{x^3} – x}}{x} = 3{x^2} – x\)
• To cancel the \(x\) out of the denominator, the first factor is an \(x\) out of the numerator. • You can cancel something only if it is multiplied by the entire numerator and denominator.	• In other words, the \(x\) in the denominator is cancelled against only one of the \(x'{\rm{s}}\) in the numerator (i.e., cancel the \(x\) only from the first term or only from the second term). • This is not a correct division. There must be an \(x\)in both terms in order to cancel.

Finding Error: Proper Use of Square Root

There appears to be a widespread misconception about the application of square roots. Students appear to be under the impression that the misconception arises because they are also asked to solve problems such as \({x^2} = 16\)

Clearly, the answer is \(x=4\), and they frequently solve by “taking the square root” of both sides. However, there is one step that is missing. Here is the correct solution method for this problem.

\({x^2} = 16\)
\(x =  \pm \sqrt {16} \)
\(x =  \pm 4\)

Example: \(\sqrt {4{x^4}} \)
Solution: \(\sqrt {4{x^4}}  = 4{x^2}\) or \(2{x^4}\), which is incorrect.
\(\sqrt {4{x^4}}  = 2{x^2}\) is the correct one.

• In the first case, students will consider the square root of either the first or the second value.
• Thus, they take the square root of \(4\) or \({x^4}\). But not all the values.
• In the correct case, they have taken the square root of all values under the root.
• Remember that, while doing the square root of monomials, consider the square root of all the values under the root.

Finding Error: Square of Expression

In general, students will find the square of only first or any other term, which leads to the error.

Example: \({\left( {3 + 2y} \right)^2}\)
Solution: \({(3 + 2y)^2} = 9 + 2{y^2}{\rm{or}}\,3 + 2{y^2}\) are incorrect calculations and lead to the error.
The correct method is \({(3 + 2y)^2} = {3^2} + {(2y)^2} + 2(3)(2y)\)
\( = {3^2} + {2^2} \cdot {y^2} + 12y\)
\( = 4{y^2} + 12y + 9\)

• Only the first value or the last value is squared in the first case.
• But, in the second case, the square is distributed to all the values.

Finding Error: Ambiguous Fractions

This is more of a notational problem than an algebra problem. In general, the ‘bad’ notation that students frequently use with fractions can lead to errors in calculation. The issue revolves around using a \({\rm{‘/’}}\) to denote a fraction, but it is a parenthesis issue. This includes fractions \(\frac{{a + b}}{{c + d}}\) such as with the use of \({\rm{‘/’}}\) to denote fractions, frequently write this fraction as \(a + b/c + d\).

These students are aware that they are recording the original fraction. However, nearly everyone else will notice the following: \(a + \frac{b}{c} + d\).

This is most certainly not the original fraction. So, if you must use \({\rm{‘/’}}\) to denote fractions, use parenthesis to indicate the numerator and denominator. As a result, you should write it as \(\frac{{(a + b)}}{{(c + d)}}\) or \((a + b/(c + d)\).

Important Points to Avoid Errors

• Coefficient \(1\) of a term is typically not displayed. However, when adding like terms, we include them in the total.
• When substituting a negative value, remember to use brackets.
• When you multiply an expression enclosed in a bracket by a constant (or variable) outside, you must multiply each term by the constant (or the variable) right outside the bracket.
• When you square a monomial, the numerical coefficient and each factor must be squared.
• Before applying any formula, make sure that the formula is truly applicable.
• When dividing a polynomial by a monomial, we divide each term in the numerator of the polynomial by the monomial in the denominator.

Solved Examples on Finding the Error

A few solved examples on finding the error has been given below:

Q.1. Find and correct the error in the given statement: \(x(3x + 2) = 3{x^2} + 2\).
Solution:
Given: \(x(3x + 2) = 3{x^2} + 2\)
We know that the given statement is incorrect, as the variable  is not distributed over the parenthesis properly.
Consider, \(a(p + q) = ap + aq\)
Similarly, \(x\left( {3x + 2} \right)\) means \(x\) needs to be multiplied separately with the terms \(3x\) and \(x\).
\( = x.3x + x.2\)
\( = 3.x.x + 2.x\)
By using the product law of exponents, \({a^m} \cdot {a^n} = {a^{m + n}}\)
\( = 3{x^{1 + 1}} + 2x\)
\( = 3{x^2} + 2x\)
Hence, the correct statement is \(x(3x + 2) = 3{x^2} + 2x\).

Q.2. Find the error and correct the statement: \(x + 2x + 3x = 5x\).
Solution:
Consider the LHS: \(x + 2x + 3x\)
We know that coefficient \(1\) of a term is usually not shown. But while adding like terms, we include it in the sum.
The sum of like terms is obtained by adding the numerical co-efficient and taking the variable common.
So, by taking  common and adding the numerical coefficients,
\(x(1 + 2 + 3) = 6x\)
Hence, the correct statement is \(x + 2x + 3x = 6x\).

Q.3. By substituting \(x =  – 3\) in \({x^2} + 5x + 4\) gives \(-15\). Check if there is an error in the above statement. If yes, correct them.
Solution:
Given:\({x^2} + 5x + 4\).
Substituting \(x=-3\) in the given expression, we get
\( = {( – 3)^2} + 5( – 3) + 4\)
\( = 9 – 15 + 4\)
\(=13-15\)
\(=-2\)
We got the value of \({x^2} + 5x + 4\) is \(-2\) by substituting \(x=-3\), But they give its is \(-15\), which is an incorrect value.
Hence, the correct answer is \(-2\).

Q.4. Preethi divided \(a+2\) by \(2\), and she wrote \(a+1\) as the answer. Her teacher Ambee said that she made an error in the calculation. Identify the error made by Preethi and correct the error.
Solution:
Preethi divided \(a+2\) by \(2\).
So, let us divide \(a+2\) by \(2\)
Here, we need to divide each term of the expression by \(2\). But, Preethi divided only one term and show wrote the answer as \(a+1\), which is incorrect.
The correct method is
\(\frac{{a + 2}}{2} = \frac{a}{2} + \frac{2}{2} = \frac{a}{2} + 1\)
Hence, the correct answer is \(\frac{a}{2} + 1\).

Q.5. Soumya wrote \({(y – 3)^2} = {y^2} + 9\). Check whether the statement written by Soumya is correct or not. If there are any errors, correct them.
Solution:
Soumya wrote \({(y – 3)^2} = {y^2} + 9\)
Here, the statement written by Soumya is incorrect. She has not distributed the square of the binomial properly. Instead, she squares the individual terms given.
The correct way of solving is given below:
\({(y – 3)^2}\)
By using the identity: \({(a – b)^2} = {a^2} – 2ab + {b^2}\)
So, \({y^2} – 2(y)(3) + {3^2}\)
\({y^2} – 6y + 9\)
So, the correct statement is \({(y – 3)^2} = {y^2} – 6y + 9\).

Summary

Several errors arise in different calculations by not using proper signs, brackets, and methods. In factorisation, these errors are more common. Dividing by a zero, improper distribution of the terms over the brackets, not proper use of parenthesis, and incorrect application of square and the square root of the expression will lead to errors.

All the errors are avoided when we look carefully and write the terms using parenthesis when required, squaring the whole values, and finding the square root of the whole values under the root. When adding the like terms, consider the \(1\) as the co-efficient of the single variable in the sum.

Frequently Asked Questions (FAQs)

The frequently asked questions on finding the error have been given below:

Q.1. How do you know when to take a negative as a factor?
Ans: According to the laws of multiplication, when a negative number is multiplied by a positive number, the product is negative. As a result, when considering a factor pair of a negative product, one of these factors must be negative, and the other must be positive.

Q.2. What is the general error while considering the coefficient of the variable?
Ans: While adding the like terms or taking the common factors, the general error neglects coefficient \(1\). The coefficient \(1\) of a term is usually not shown. But while adding like terms, we include it in the sum.

Q.3. What is the error to be made in dividing the polynomial by the monomial?
Ans: The general error made in dividing the polynomial by the monomial is not dividing each term by the monomial. When dividing a polynomial by a monomial, we divide each term in the numerator of the polynomial by the monomial in the denominator.

Q.4. What needs to be remembered to avoid while squaring the monomials?
Ans: When you square a monomial, you must square both the numerical coefficient and each factor.
Example: \({(2x)^2} = {2^2} \cdot {x^2} = 4{x^2}\)

Q.5. What should be kept in mind while multiplying the expression enclosed in brackets?
Ans: Remember that when you multiply the expression enclosed in a bracket by a constant (or variable) outside, you must multiply each term of the expression by the constant (or the variable).

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