Monomial Factors

Example:

Identify all of the common monomial factors in

25x ⁴y ³zw ² + 150x ³y ³zr ⁴ + 30x ²y ³zw

solution:

Our goal 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.

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 ⁴y ³zw 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 and w.

(Note that if we had chosen to key our analysis to the third term, the 30x ²y ³zw, our list of simple factors would be 2, 3, 5, x, y, z, w – slightly longer. 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, and 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 ⁴y ³zw ² contains 5 ²

150x ³y ³zr ⁴ contains 5 ²

30x ²y ³zw 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 ⁴y ³zw ² contains x ⁴

150x ³y ³zr ⁴ contains x ³

30x ²y ³zw contains x ²

Thus, the highest power of x common to all three terms is x ², and so x ² is a common monomial factor of the entire expression.

(iii) y:

25x ⁴y ³zw ² contains y ³

150x ³y ³zr ⁴ contains y ³

30x ²y ³zw containsy ³

Thus, the highest power of y common to all three terms is y ³, and so y ³ is a common monomial factor of the entire expression.

(iv) z:

25x ⁴y ³zw ² contains z ¹

150x ³y ³zr ⁴ contains z ¹

30x ²y ³zw contains z ¹

Thus, the highest power of z common to all three terms is z ¹ = z, and so z is a common monomial factor of the entire expression.

(v) w:

25x ⁴y ³zw ² contains w ²

150x ³y ³zr ⁴ does not contain w (or contains w ⁰)

30x ²y ³zw 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, and 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, and z. Our strategy has guaranteed that the product of these, 5x ²y ³z 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 ⁴y ³zw ² + 150x ³y ³zr ⁴ + 30x ²y ³zw = (5x ²y ³z)( whatever else)

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 ⁴y ³zw ² = 5 2x ⁴y 3zw ² = (5x ²y ³z)(5x ²w ²)

since to get 25x ⁴y ³zw ² = 5 2x ⁴y ³zw ² from 5x ²y ³z, we need an additional factor of 5 (to make the 5 ²), an additional factor of x 2 (to make the x 4), and 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 ³y ³zr ⁴ = 2 · 3 · 5 ² x ³y ³zr ⁴ = (5x ²y ³z)(2 · 3 · 5xr ⁴) = (5x ²y ³z)(30xr ⁴)

and

30x ²y ³zw = 2 · 3 · 5x ²y ³zw = (5x ²y ³z)(2 · 3w) = (5x ²y ³z)(6w)

So, 25x ⁴y ³zw ² + 150x ³y ³zr ⁴ + 30x ²y ³zw = (5x ²y ³z)(5x ²w ² + 30xr 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-hand side of this result gives precisely the expression on the left-hand side, confirming that the two forms are mathematically equivalent.