August 28th August 28th Monomial FactorsThe 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. solution: We see that 2xy = 2 · x · y and 6yz = 2 · 3 · y · z We can quickly recognize that both terms contain the factors 2 and y in common. Thus 2xy + 6yz = 2y(x + 3z). The right hand 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 remultiplication to confirm that the original expression is regenerated.
Example 2: Remove all common monomial factors from: 16x^{ 3} – 24x^{ 2} + 84x. solution: 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 easytosee simple factorization really is complete and 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: 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, and 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} = 2 ^{3 }· 3 · x^{ 2} and 84x = 2^{ 2} · 3 · 7 · 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
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^{ 2} + 84x = 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, and so all of the common monomial factors have been “factoredout” 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.

Copyright © 20072015, algebraonline.com. All rights reserved.