ALGEBRAIC EXPRESSIONS
An algebraic expression is a mathematical expression that consists of variables, numbers and algebraic operations. Let us define some words that we will use in this topic:
- Variable – A letter or symbol that is used to represent a number.
- Constant – A number that has a fixed value and stands on its own, such as the number 5 or 8. A constant can also be represented by a letter such as A, B, or C, provided that the letter stands in for a fixed value.
- Coefficient – A number that is multiplied by a variable. It is written before the variable.
- Term – A single part of an expression, separated by an addition or subtraction symbol.
Algebraic expressions are fundamental in solving practical problems in almost all fields, from science and engineering to finance and medicine.
Simplification of Algebraic Expressions
Simplifying algebraic expressions involves removing adding or subtracting all the like terms to remain with only unlike terms.
Like terms are those that have the same variable(s) to the same power. For example, 3a and 5a are like terms, but 3a and 3a^2 are unlike terms. When simplifying like terms, we add or subtract the coefficients while the variables remain the same.
Remember that the order in which numbers or variables are multiplied does not matter. e.g. a^2b and ba^2 are equal.
Examples
Simplify the following expressions:
(a) \displaystyle-\frac{r}{9}+\frac{5g}{6}-\frac{r}{3}
We need to collect like terms first. Collecting like terms is rearranging the terms in the expression such that the like terms are next to one another. Here, \frac{r}{9} and \frac{r}{3} are like terms. When we put them next to each other, we get: \frac{5g}{6}-\frac{r}{9}-\frac{r}{3} Note that the sign before each term remains the same even after rearranging. To simplify the like terms, we find the LCM of the denominators 3 and 9, which is 9. We can now rewrite the expression as: \frac{5g}{6}-\frac{r}{9}-\frac{3r}{9} We are subtracting both \dfrac{r}{9} and \dfrac{3r}{9} in this expression. This is essentially the same thing as subtracting \dfrac{r}{9}+\dfrac{3r}{9}, which is \dfrac{4r}{9}. The final, simplified expression is therefore \displaystyle\boxed{ \frac{5g}{6}-\frac{4r}{9} } .
(b) jkl+klj+lkj+2jk
jkl, klj and lkj are like terms, since they are just the same variables multiplied in different orders. jkl+klj+lkj=3jkl Therefore, our simplified expression is \boxed{3jkl+2jk} .
Expansion
Expansion in algebra is removing all the brackets in an expression by multiplying the terms in the brackets by the factor outside the brackets. Each term is multiplied by the factor. It is easy to make a mistake and only multiply the first term, so ensure that you multiply the factor by each and every term in the brackets.
Let us consider how we would expand the expression 5(x+3). We first multiply the 5 by each term. This is called distributing the factor.
\Large 5\times x + 5\times3 \Large=5x+15
Examples
Expand the following algebraic expressions:
(a) \displaystyle\frac{2}{y} \left( \frac{1}{2}y+y^2+y^3 \right)
We need to distribute the \displaystyle\frac{2}{y}. \frac{2}{y}\times\frac{1}{2}y+\frac{2}{y}\times y^2+\frac{2}{y}\times y^3 = 1+2y+2y^2
(b) y^2-3x(xy-x^2)
You may be tempted to simply multiply each term in the brackets by 3x, but remember that the sign before 3x is negative. So, we need to multiply each term by -3x. y^2+(-3x)\times xy-(-3x)\times x^2 y^2 -3x^2y – (-3x^3) \boxed{y^2 – 3x^2y +3x^3}
Factorisation
Factorisation is the reverse of expansion. When factorising, we introduce brackets to the expression. This can be useful in solving more complicated algebraic expressions, which we will see as we advance in math.
To factorise an expression, we take out the GCD of all the terms, then divide each term by the GCD to find the new terms in the brackets.
Examples
Factorise the following expressions:
(a) 4x^2+6x
The GCD of 4x^2 and 6x is 2x as 2 is the biggest number that divides 4 and 6 and x is the largest variable that divides x and x^2. We put this outside the bracket:
2x(?+?)
We find the missing terms in the brackets by dividing each term by the HCF:
4x^2\div2x=2x and 6x\div2x=3
This gives: 2x(2x+3)
To check if our answer is correct, we can expand it and see if the result is the same as the original expression. Expand it yourself to verify the answer.
(b) 4x^2y+20x^4y^2-36x^3y
Here, 4x^2y is the common factor (go through the steps to see how we found it). We now take this factor out of each term to get the desired factorised expression: 4x^2y(1+5x^2y-9x)
Factoring by Grouping
The best way to understand this concept is by looking at an example. We will factor the expression: ax+a+b+bx
There is no common factor to all the terms in the expression. However, if we group the first two terms together and the last two terms together, each group has its own HCF. (ax+a)+(b+bx) The first group has a HCF of a, while the second one has a HCF of b. We can factor these out to obtain the following expression:
a(x+1)+b(x+1)
Notice that this reveals another common factor between the two terms: (x+1). We can factor out this common factor.
(x+1)(a+b)
And we’re done! We just factored an expression that originally seemed to have no common factors. This method is called factorisation by grouping. This method is used in solving quadratic equations which we will work with in Form 2.
Let us factorise another expression, 3x^2-6x-4x+8.
It does not have a common factor to all the terms, so we will need to factorise by grouping. The first step is to arrange the terms such that the first two terms and the last two terms have common factors. Luckily, they are already in the correct order so we can skip this step. We can group this expression as follows:
(3x^2-6x)+(-4x+8)
Note a very crucial thing we just did, a “+” sign was added between the groupings. This is because the third term (-4x) is negative, and the sign of the term must be included in the grouping.
A common error is to group 3x^2-6x-4x+8 as (3x^2-6x)-(4x+8). This grouping, however, simplifies to 3x^2-6x-4x\textcolor{red}{-8} , which is not the same as the original expression. Try expanding the incorrect grouping to see for yourself that this would happen. This is a very common mistake, so keep an eye out for this situation.
The first group has a HCF of 3x, and the second has a HCF of -4. Factoring these out, we get:
3x(x-2)+(-4)(x-2)
= 3x(x-2)-4(x-2)
Note that we factor out -4 instead of 4 in order to get the common factor x-2. If we factored out 4, the factor in the second term would be -x+2.
Now that we have the common factor x-2, we can factor it out to get the final expression: (x-2)(3x-4)
Algebraic Fractions
In algebra, fractions can be added and subtracted by finding the LCM of the denominators. This works the same way as in arithmetic.
Example 1
Express the following as a single fraction:
\frac{a+b}{a}-\frac{b-a}{b}
We first need to find the LCM of the denominators. The LCM of a and b is ab. So, this will be our new denominator.
\frac{?(a+b)}{ab}-\frac{?(b-a)}{ab}
We need to find what values will make the fractions have the same values as the original, because we need to maintain the original definition of the expression. These values would be b and a respectively. \frac{b(a+b)}{ab}-\frac{a(b-a)}{ab} Combining this into a single fraction, we get: \frac{b(a+b)-a(b-a)}{ab} We can simplify this expression further by expanding it.\frac{ba+b^2-ab+a^2}{ab}\frac{a^2+b^2}{ab}
One of the main uses of factorisation is simplifying expressions. Let us see some examples.
Example 2
Simplify the following expression: \displaystyle\frac{18ar-18am}{9am-9ar}
The first thing we need to do is look for any common factors in the numerator and common factors in the denominator. Here, 18a is common in the numerator, and 9a is common in the denominator. Let us take out these factors.
\frac{18a(r-m)}{9a(m-r)} \frac{2\>\>\cancel{18a}(r-m)}{\cancel{9a}(m-r)} \frac{2(r-m)}{(m-r)} This expression can still be simplified further. Whenever two variables are being subtracted but in different orders in the numerator and denominator, you can always flip the subtraction by factoring out -1. Let us try taking out -1 in the denominator. \frac{2(r-m)}{-(-m+r)} \frac{2(r-m)}{-(r-m)} \frac{2\cancel{(r-m)}}{-\cancel{(r-m)}} \frac{2}{-1}=-2
Example 3
Simplify the expression: \displaystyle \frac{x^2+x-4xy-4y}{(x+1)(4y^2-xy)}
The goal is always to have common factors in the numerator and denominator when it comes to simplification question. So, let us try to factor the numerator. We can factor it by grouping: x^2+x-4xy-4y x(x+1)-4y(x+1) (x+1)(x-4y) Our fraction is now: \frac{(x+1)(x-4y)}{(x+1)(4y^2-xy)} \frac{\cancel{(x+1)}(x-4y)}{\cancel{(x+1)}(4y^2-xy)} \frac{x-4y}{4y^2-xy} We can further simplify the fraction by factoring the denominator: \frac{-1(4y-x)}{y(4y-x)} \frac{-1\cancel{(4y-x)}}{y\cancel{(4y-x)}} =-\frac{1}{y}