Monomial Factors. Sometimes the expression to be factored is simple enough to be able to use straightforward inspection.

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1 The Mathematics 11 Competency Test Monomial Factors The first stage of factoring an algebraic expression involves the identification of any factors which are monomials. We will describe the process by working through several examples in detail. Sometimes the expression to be factored is simple enough to be able to use straightforward inspection. Example 1: Factor the expression: 2xy + 6yz. We see that 2xy = 2 x y 6yz = 2 3 y z We can quickly recognize that both terms contain the factors 2 y in common. Thus 2xy + 6yz = 2y(x + 3z). The right h side here is in factored form because it is a single term only. As it is written, it does not consist of two or more parts which are connected by plus or minus signs. It is just a single term which is the product of three factors in this case. That the factored expression above is mathematically equivalent to the original expression is easily verified by multiplication. 2y(x + 3z) = (2y)(x) + (2y)(3z) = 2yx + 6yz = 2xy + 6yz. In fact, you can always verify that a factorization is correct by re-multiplication to confirm that the original expression is regenerated. Example 2: Remove all common monomial factors from: 16x 3 24x x. This expression is a bit more complicated than the one in the previous example. With it, we can begin to implement the systematic method that should be used in all but the very simplest cases where the factorization is immediately obvious. (In fact, you should probably always do the systematic method, because that is really the only way to make sure that what you thought was an easy-to-see simple factorization really is complete correct!) We start by selecting any term in the expression -- it doesn t matter which one we pick, so we will pick the first one here: 16x 3. Enumerate the factors which make up this term: David W. Sabo (2003) Monomial Factors Page 1 of 7

2 16x 3 = 2 4 x 3 This means that any monomial factor common to all three terms in the original expression must either be a power of 2 or a power of x. This is because monomial factors common to all three terms must be factors of each individual term, hence they must be a factor of the first term (or whichever one you picked). So, all we have to do is check what power of 2 is common to all three terms, what power of x is common to all three terms. It s probably easiest at this stage to write out the factorizations of each term for easy reference: 16x 3 = 2 4 x 3 24x 2 = x 2 84x = x 1 (The method for resolving whole numbers into products of prime factors has been described in an earlier document in these notes.) From this we see that all three terms contain a factor of 2 2, so this is a factor of the entire expression all three terms contain a factor of x 1 or x, so this is a factor of the entire expression the above two bullets in this list cover all possible factors of the first term, so cover all possible monomial factors of the entire expression. Thus, 2 2 x = 4x is the most extensive monomial factor common to all three terms of the original expression. We can write 16x 3 24x x = 4x(4x 2 6x + 21) The trinomial in brackets is obtained by removing the factor 4x from each of the three terms in the original expression: 16x 3 4x = 4x 2-24x 2 4x = -6x 84x 4x = 21 The expression on the right above is the required factored expression here. The terms of the trinomial in the brackets contain no further common monomial factors, so all of the common monomial factors have been factored-out as required. You should take a minute here to verify that if you multiply to remove the brackets from the factored expression, you get precisely the original expression back again. Example 3: Remove all common monomial factors from 42b 2 y 28by 2. This example is very similar to the previous one, so you should try it as a practice problem on your own first, before looking at our brief outline of a solution. First, write out the factorization of each of the two terms explicitly: David W. Sabo (2003) Monomial Factors Page 2 of 7

3 42b 2 y = b 2 y 1 28by 2 = b 1 y 2 Notice that we ve written exponents of symbols explicitly, even if those exponents are 1 (just as a visual cue when we check now for common factors between the two terms). It is also helpful to sort the factors of the individual terms in a common order. Here numerical prime factors are sorted from smallest to largest (going left to right) symbolic factors are sorted alphabetically. Comparison of these two factorizations indicates immediately that the common factors are 2, 7, b, y, all to the first power. Thus 42b 2 y 28by 2 = 2 7 b y(3b 2y) = 14by(3b 2y). The terms in the expression in brackets on the right here are obtained by taking what s left of each of the original terms when the common factors are removed. Thus, since 42b 2 y = b 2 y 1 when we remove the factors 2, 7, b 1, y 1, all that s left is the 3 one of the factors b, or 3b. A similar inspection indicates that after removal of these four common factors from 28by 2, all that is left is the factors 2 y, each to the first power. Thus, the required factorization here is 42b 2 y 28by 2 =14by(3b 2y). We ll leave it up to you to verify that this is correct by multiplication. Example 4: Identify all of the common monomial factors in 25x 4 y 3 zw x 3 y 3 zr x 2 y 3 zw This looks bad (somewhat intentional, to demonstrate it is as easily hled using a systematic approach as any of the previous examples). The goal right here is to rewrite this trinomial in the form: (a monomial) (whatever else) Note that this pattern is a single term which is the product of two parts. The a monomial part will be identified here, in the process determining what the whatever else part is. (Before leaving the topic of factoring, we will also demonstrate things to check for factoring the whatever else part a bit further in some cases.) In solving this example, we ll use a slight variation on the methods demonstrated for the previous two examples. We start by selecting any of the three terms to guide the process it doesn t matter which term you choose, though there may be a slight savings in work if a simpler-looking term is used. So, we decide to base this analysis on the parts of the first term, the 25x 4 y 3 zw 2. Next, identify all of the simple factors of the selected term, including prime factors in the numerical coefficient. The term we have chosen to work from has simple factors 5, x, y, z w. David W. Sabo (2003) Monomial Factors Page 3 of 7

4 (Note that if we had chosen to key our analysis to the third term, the 30x 2 y 3 zw, our list of simple factors would be 2, 3, 5, x, y, z, w slightly longer. Usually there s very little reason to prefer one term over any other, so don t spend a lot of time making your selection here.) Now, go through the list of these simple factors in your key term one at a time, determine the highest power to which each of them occurs in all (three, in this case) terms of the entire expression. This highest power will then be a common monomial factor of the entire expression. (Of course, if an item in this list doesn t occur at all in one of the terms, then this highest power will be the zero power that item is not a common factor of all of the terms in the expression.) So, for our example, we have five items in the list of potential common monomial factors to check. (i) 5: 25x 4 y 3 zw 2 contains x 3 y 3 zr 4 contains x 2 y 3 zw contains 5 1 The highest power of 5 common to all three terms (which is actually the lowest power that occurs in this list) is 5 1 = 5. Thus, 5 is a common factor of all three terms. (ii) x: 25x 4 y 3 zw 2 contains x 4 150x 3 y 3 zr 4 contains x 3 30x 2 y 3 zw contains x 2 Thus, the highest power of x common to all three terms is x 2, so x 2 is a common monomial factor of the entire expression. (iii) y: 25x 4 y 3 zw 2 contains y 3 150x 3 y 3 zr 4 contains y 3 30x 2 y 3 zw contains y 3 Thus, the highest power of y common to all three terms is y 3, so y 3 is a common monomial factor of the entire expression. (iv) z: 25x 4 y 3 zw 2 contains z 1 150x 3 y 3 zr 4 contains z 1 30x 2 y 3 zw contains z 1 Thus, the highest power of z common to all three terms is z 1 = z, so z is a common monomial factor of the entire expression. (v) w: 25x 4 y 3 zw 2 contains w 2 150x 3 y 3 zr 4 does not contain w (or contains w 0 ) 30x 2 y 3 zw contains w 1 Thus the highest power of w common to all three terms is w 0 = 1. This is the same thing as saying that there is no power of w common to all three terms, so there is no power of w which is a common monomial factor of all three terms. So, the common monomial factors of the entire three-term expression have been identified as 5, x 2, y 3, z. Our strategy has guaranteed that the product of these, 5x 2 y 3 z makes up the greatest monomial factor common to all three terms. Thus, going back to the template of our original goal, we can now write that 25x 4 y 3 zw x 3 y 3 zr x 2 y 3 zw = (5x 2 y 3 z)(whatever else) David W. Sabo (2003) Monomial Factors Page 4 of 7

5 All that needs determining yet is the form of the whatever else part. This we do in the same way as was done in the previous examples. Identify what is left of each term after the common monomial factor is removed. For the first term, 25x 4 y 3 zw 2 = 5 2 x 4 y 3 zw 2 = (5x 2 y 3 z)(5x 2 w 2 ) since to get 25x 4 y 3 zw 2 = 5 2 x 4 y 3 zw 2 from 5x 2 y 3 z, we need an additional factor of 5 (to make the 5 2 ), an additional factor of x 2 (to make the x 4 ), an additional factor of w 2 (to make the w 2 ). Multiplying the two factors in brackets on the right above is seen to regenerate the original term on the left. Now, repeat this process with each of the remaining two terms: 150x 3 y 3 zr 4 = x 3 y 3 zr 4 = (5x 2 y 3 z)(2 3 5xr 4 ) = (5x 2 y 3 z)(30xr 4 ) 30x 2 y 3 zw = 2 3 5x 2 y 3 zw = (5x 2 y 3 z)(2 3w) = (5x 2 y 3 z)(6w) So, 25x 4 y 3 zw x 3 y 3 zr x 2 y 3 zw = (5x 2 y 3 z)(5x 2 w xr 4 + 6w) This completes the operation of identifying the common monomial factors in the original expression. You should verify that multiplying to remove the brackets on the right-h side of this result gives precisely the expression on the left-h side, confirming that the two forms are mathematically equivalent. We have the required answer, but a couple more observations about this example may help you solve other problems of this sort. Notice that the second term in the example contained a power, r 4, of r. Yet our detailed analysis of potential monomial factors did not consider powers of r at all. We didn t consider powers of r, because r did not occur in the term 25x 4 y 3 zw 2 on which we were basing our analysis. But since r did not occur in this term at all, a power of r could not possibly be a common factor of all three terms. Because any common monomial factors must be a factor of the term on which we base our analysis, any symbols which do not appear in that term need not be considered at all. Hence the value in keying the analysis on a simpler-looking rather than more complicated-looking term in the original expression. Once you underst the strategy of the method displayed in great detail above, you may be able to accomplish the same end with much less writing without sacrificing the systematic approach. You still key the analysis on one of the terms in the expression, but common powers of factors are identified removed from the original expression in a stepwise fashion. 25x 4 y 3 zw x 3 y 3 zr x 2 y 3 zw 5 is a factor of the first term, each of the remaining terms contain a factor of 5, so remove a factor of 5 from all three terms. = 5(5x 4 y 3 zw x 3 y 3 zr 4 + 6x 2 y 3 zw) The only remaining numerical factor in the first term in the brackets is 5, but not all of the remaining terms in brackets contain a factor of 5, so there are no more numerical factors to remove. The first symbolic factor is a power of x. All three terms have a factor of at least x 2, so remove a factor of x 2 from the bracketed expression. David W. Sabo (2003) Monomial Factors Page 5 of 7

6 =5x 2 (5x 2 y 3 zw xy 3 zr 4 + 6y 3 zw) The next symbolic factor in the first term in the brackets is a power of y. All three terms contain a factor of y 3, so remove it. =5x 2 y 3 (5x 2 zw xzr 4 + 6zw) The next symbolic factor in the first term in the brackets is a z. Each of the three terms in brackets contains a factor z, so remove it. =5x 2 y 3 z(5x 2 w xr 4 + 6w) The next symbolic factor in the first term in brackets is a power of w. However not all of the remaining terms in the brackets contain a factor which is a power of w, so there is no power of w to be removed from the bracketed expression. Since we have run out of factors in the first term in brackets, we are done. The resultant expression has all common monomial factors identified. Notice that at each step, we needed to deal only with the expression left in the brackets, which is also generally getting simpler with each step (at least with those steps in which we ve been able to remove a factor). You can also check your work at each step because if you remultiply by the most recently removed factor, you must obtain the expression from the previous step. This has been a rather lengthy discussion of issues approaches to factoring one fairly complicated expression. To conclude this document, we ll briefly demonstrate the identification of monomial factors for a couple of additional expressions. Example 5: Remove all common monomial factors from the expression 3r 4 s 3 + 6r 3 s 4 105r 2 s 5 The first term in this expression has factors of 3, r, s. So, we need to check whether 3 is a common factor in all three terms, whether there are common powers of r s in all three terms. 3r 4 s 3 + 6r 3 s 4 105r 2 s 5 3 is a factor of the first term, upon inspection is found to be a factor of the remaining two terms as well. So, remove a factor of 3 from all three terms. = 3(r 4 s 3 + 2r 3 s 4 35r 2 s 5 ) Each term contains a factor of r 2. Remove it. = 3r 2 (r 2 s 3 + 2rs 4 35s 5 ) Each term in the brackets now contains a factor of s 3. Remove it. = 3r 2 s 3 (r 2 + 2rs 35s 2 ) Since the first term in brackets contains only powers of r, but the third term contains no factor of r, there cannot be any further common monomial factors in the three terms of the expression in brackets. The required factorization of monomial factors is thus complete. Thus, with all fo the common monomial factors identified, we can write the final result as: 3r 4 s 3 + 6r 3 s 4 105r 2 s 5 = 3r 2 s 3 (r 2 + 2rs 35s 2 ) David W. Sabo (2003) Monomial Factors Page 6 of 7

7 (Once you ve read the next section in these notes, you ll be able to see that the expression in brackets in the final answer of this last example can, in fact, be written as a product, though not of monomial factors. So, this is a final answer only as far as the methods presented so far here are concerned.) Example 6: Remove all common monomial factors from the expression 54x 3 y 2 z 150xy 4 z Looking at the first term here, we see that in addition to a possible common numerical factor, we need to check for common powers of x, y, z in the two terms. The hardest part of this example is in determining the common numerical factor in the two terms. However, writing the two numerical coefficients as products of prime factors (we described how to do this in the note called Prime Factors in the section on numerical fractions), we get 54 = 2 x 3 x 3 x = 2 x 3 x 5 x 5 Thus, both terms share a factor of 2 x 3 = 6. Proceeding as in previous examples, we now get: 54x 3 y 2 z 150xy 4 z = 6(9x 3 y 2 z 25xy 4 z) removing the already identified common numerical factor of 6. Now we note that both terms share a factor of x 1 = x, so remove this factor. = 6x(9x 2 y 2 z 25y 4 z) Next, there is a common factor of y 2 shared by the two terms, so remove it. = 6xy 2 (9x 2 z 25y 2 z) Finally, both terms share a factor of z, so remove it. = 6xy 2 z(9x 2 25y 2 ) Thus, the final answer here is 54x 3 y 2 z 150xy 4 z = 6xy 2 z(9x 2 25y 2 ) Shortly, we ll demonstrate that the bracketed expression in the answer given above in Example 6 can be factored into the product of two binomials, so that the most factored form of the original expression is 54x 3 y 2 z 150xy 4 z = 6xy 2 z(3x + 5y)(3x 5y). However, the identification of all common monomial factors is the necessary first step before more specialized types of factorization can be attempted. David W. Sabo (2003) Monomial Factors Page 7 of 7