Factorization and quadratic expressions

Factorization and quadratic expressions

Factorizing is the process of finding two or more factors which when multiplied together will result in the given expression. Therefore, factorizing is really the opposite of multiplication or finding the product. It was for this reason that we first considered the simpler process of finding the product.

Thus, for example, x(y + z) = xy + xz. This product resulted from the multiplication of the two factors x and (y + z). If we now unpick the product you should be able to see that x is a common factor that appears in both terms of the product.

What about the expression x² – 16? I hope you are able to recognize the fact that this expression is an example of the difference between two squares. Therefore, we can write down the factors immediately as (x + 4) and (x – 4), look back at Example 4.8, if you are unsure. We can check the validity of our factors by multiplying and checking that the product we get is identical to the original expression that we were required to factorize, that is (x + 4)(x – 4) = x² – 4x + 4x– 16 = x² – 16, as required.

Here is an illustration of the technique for finding common factors that is very useful when simplifying algebraic expressions, Example 4.10, illustrates the method that was explained earlier.

Example 4.10

Simplify the following expressions by extracting common factors:

(a) abc + bcd                  and           (b) wx²yz – w²xy²z + wxyz².

(a) Looking at the two terms in the expression, it can be seen that be is common to both; therefore, it can be removed from both the terms. After removal, it is necessary to bracket what is left of the original expression, so that the factors are held together by the arithmetic operation of multiplication. So, we get bc(a + d) and we know that these are factors because if we multiply these factors back together we get bco + bcd and following convention this may be written in alphabetical order as obc + bcd, as before.

(b) A similar argument follows for this example, it is just a question of recognizing what is common to all three terms in the expression. I hope you can see that wxyz is common to all three terms and because x² = x – x (x multiplied by x) and y² = y – y, etc., then on removal of the common factor we are left with the factors wxyz(x – wy + z). Again if these two sets of factors are multiplied up we get back to our original expression.

Quadratic expressions involve at least one term that has an element in it raised to the power two, that is squared. Suppose you are asked to factorize the expression a² – 6a + 9, how do we go about it. Well a good place to start is with the term involving the highest power of the variable, that is a². Remember that convention dictates we lay out our expression in descending powers of the unknown, starting with the highest power positioned at the extreme left side of the expression.

a² can only have factors of itself and 1 or a and a, therefore, ignoring the trivial factors, a² = a – a. At the other end of the expression we have the natural number 9, this has the trivial factors 1 and 9 or the factors 3 and 3 or –3 and –3. Note the importance of considering the negative case, where from the laws of signs (–3)(–3) = 9. So, now, we have several sets of factors we can try, these are:

(i) (a + 3)(a + 3)       or    (ii) (a – 3)(a – 3) or    (iii) (a + 3)(a – 3). Now, we could try multiplying up each set of factors until we obtained the required result, that is determine the factors by trial and error. This does become rather tedious when there are a significant number of possibilities. So, before resorting to this method, we need to see if we can eliminate some combinations of factors, by applying one or two simple rules.

I hope you can see why the factors (a + 3)(a – 3) can be immediately excluded. These are the factors for the difference between two squares, which is not the original expression we need to factorize.

What about the factors (a + 3)(a + 3), both factors contain only positive terms, therefore any of their products must also be positive by the laws of signs! In our expression a² – 6a + 9 there is a minus sign, so again this set of factors may be eliminated. This leaves us with the factors

(a – 3)(a – 3) and on multiplication we get: (a – 3)(a – 3) = a² – 3a – 3a + 9 = a² – 6a + 9, giving us the correct result.

You may have noticed that we left out the sets of factors (a – 1)(a – 9), (a – 1)(a + 9), (a + 1)(a – 9) and (a + 1)(a + 9) from our original group of possibles! Well in the case of (a + 1)(a + 9), this would be eliminated using the laws of signs, but what about the rest?

There is one more very useful technique we can employ when considering just two factors. This technique enables us to check the accuracy of our factors by determining the middle term of the expression we are required to factorize. So, in our case for the expression 2a – 6a + 9, –6a is the middle term.

The middle term is derived from our chosen factors by multiplying the outer terms, multiplying the inner terms and adding.

So, in the case of the correct factors (a – 3)(a – 3), the outer terms are a and –3, which on multiplication (a)(–3) = –3a and similarly the inner terms (–3)(a) = –3a and so their sum = –3a + (–3a) = –6a, as required.

If we try this technique to any of the above factors involving l and 9, we will see that they can be quickly eliminated. For example (a – 1) (a – 9) has an outer product of (a) (–9) = –9a and an inner product of (–1) (a) = –a, which when                          added      = – 9a – a = –10a, this of course is incorrect.