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Factoring: Special Cases Part II - Problem 2
When I’m asked to do factoring by grouping or when if I’m given a polynomial that has four terms I really want to make sure it’s in standard form meaning the exponents are in decreasing order. So if you check it out these exponents are in decreasing order, I’m ready to start.
The way factoring by grouping works is first we cut it into two different halves and then we are going to factor each half individually. Here's what I mean I’m looking for what monomial multiplies into both if these terms. What number goes into 12 and 9? Well 3 and then how many Xs go into x to the third and also x²? That would be x² so now I’m like un-distributing. 3 times what gives me 12, 4. X² times what gives me x to the third? That's going to be x.
Then here I have to think 3 times what gives me 9, that's -3 and then x² times 1 gives me the x². Okay so this first half is done it’s factored let's look at the second half. I want to look for what monomial goes into both of these. What I notice right away that 5 goes into 20 and 15 neither 15. So I’m going to factor that out what's going to be left is going to be 4x minus 3.
Notice anything interesting? These are the same binomials that's my greatest common factor so what I can do to write my final factored form of the original four term polynomial is have the 4x minus 3 piece multiplied by 3x² plus 5.
That's my final answer the way to check it would be to FOIL I’ll do that really quickly at least you guys can trust me. 12x to the third is my first outers I will have plus 20x inners I’ll have take away 9x squared and then last is take away 15 and if you look this here result is the exact same as my original problem only these two terms are switch but you guys can see that by doing these factoring by grouping process where I look at this group first then that group second, look for the common binomial, that's how I can get the product that represents this original polynomial it’s tricky but you'll get the hang of it once you do some homework problems.