Algebra Tutorials! Saturday 16th of February

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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Monomial Factors

If we were asked to identify all of the common monomial factors in

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

we would reach the answer:

25x 4y 3zw 2 + 150x 3y 3zr 4 + 30x 2y 3zw = (5x 2y 3z)(5x 2w 2 + 30xr 4 + 6w).

However, 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 4y 3zw 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 understand 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 and removed from the original expression in a stepwise fashion.

 25x 4y 3zw 2 + 150x 3y 3zr 4 + 30x 2y 3zw 5 is a factor of the first term, and each of the remaining terms contain a factor of 5, so remove a factor of 5 from all three terms. = 5(5x 4y 3zw 2 + 30x 3y 3zr 4 + 6x 2y 3zw) 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. =5x 2(5x 2y 3zw 2 + 30xy 3zr 4 + 6y 3zw) 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 2y 3(5x 2zw 2 + 30xzr 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 2y 3z(5x 2w 2 + 30xr 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 and 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.