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1 250) Factoring Polynomials Sometimes when we try to solve or simplify an equation or expression involving polynomials the way that it looks can hinder our progress in finding a solution. Factorization is a means of rewriting an expression in a way that can make it much easier to work with and simplify. 1 GCF We begin with the easiest and often the most overlooked step in factoring: finding a greatest common factor gcf). Take for example the following binomial: 6x + 26 Notice that the coefficient on x and the constant term are both multiples of 2. We can use this fact to simplify the expression 6x )x ) 2 3x) + 213) 2 3x + 13) Now sometimes the common factor includes variables, for example 1.1) 121x 2 y + 33y 2 11y11x 2 + 3y) 1.2) Looking for a GCF is always the first thing one should do when factoring. 2 2Term The 2term formulae are very simple to memorize, the difficulty is that they can be buried in the expression. For this reason we will do little more than state them. 2.1 Difference of Squares The difference of squares is an easy pattern to memorize: A very simple example may look much like this: 2.2 Difference of Cubes 2.3 Sum of Cubes 2.4 Conclusion x 2 y 2 x y)x + y) 2.1) x 2 4 x 2)x + 2) 2.2) x 3 y 3 x y)x 2 + xy + y 2 ) 2.3) x 3 + y 3 x + y)x 2 xy + y 2 ) 2.4) As easy as these patterns are, sometimes it can be hidden in much stranger looking expressions like the following which can require quite a bit of algebraic manipulation. 8x xy x x x y 6 2 x4x 2 9y 3 ) 2 ) 2 x2x) 2 3y 3 ) 2 ) 2 x2x 3y 3 )2x + 3y 3 ) 2.5) 1 Last Updated: June 27, 2012
2 250) term Leading Coefficient Now we will turn our attention to factoring trinomials of the form To do this we simply 1. Find two numbers p and q such that p + q b and pq c 2. Write x 2 + bx + c x + p)x + q) x 2 + bx + c 3.1) This could be rather complicated but since we are usually talking about the case where b and c are integers, and we are usually only willing to factor if we can find integers p and q, it really isn t too bad. In such a case to reduce the number of guesses required to find such a p and q, we start by determining the pairs of factors of c and then from there we check the sums to find which pair if any) sums to b. For example: Now we look at the pairs of factors for 6: x 2 x 6 3.2) 1, 6), 1, 6), 2, 3), 2, 3) Next do the sums 5, 5, 1, 1 Since 1 b we determine that the desired p, q) is 2,3) and hence x 2 x 6 x 3)x + 2) 3.3) 3.2 Non1 Leading Coefficient Naturally the next step is to generalize the above procedure to trinomials of the form ax 2 + bx + c where a 0 3.4) If a 1 then we just have the case covered in section 3.1) The first question we should ask ourselves is can we factor out a? Remember always look for a GCF before doing anything else! Factoring out the Offending Coefficient In the case where a is a factor of both b and c, that is where b aj and c ak for some integers j and k then we can reduce it to the previous case by way of ax 2 + bx + c ax 2 + aj)x + ak) ax 2 + jx + k) 3.5) This might look a little strange, so a concrete, simple example is helpful: 5x x x 2 + 5x + 6) 5x + 3)x + 2) 3.6) 2 Last Updated: June 27, 2012
3 250) Fairly) General Trinomial Factoring If we can t simply factor out the leading coefficient, we must work on a slightly more complicated task to factor ax 2 + bx + c: Let s factor the following trinomial: 6x x + 35 The first step is to find a pair of numbers p, q) such that: pq ac p + q b So in our example we need to find two numbers whose product is and whose sum is 29. Let s try a few pairs of factors for 210: but So that pair is no good but So that pair is not good either and So 14,15) is our pair. Now we use our two numbers to decompose the middle term, group and factor. This sounds a little vague, but working this example will make things clear. 6x x x x + 15x) + 35 Decompose the middle term 6x x) + 15x + 35) Regroup terms 2x)3x + 7) + 53x + 7) Factor the groups 3x + 7)2x + 5) Factor out the common binomial factor 6x x x + 7)2x + 5) With this procedure the hardest part is finding our pair of numbers. Luckily finding such numbers gets quicker with practice. 3 Last Updated: June 27, 2012
4 250) Perfect Square Trinomial Given a trinomial of the form then Example: a 2 x 2 + bx + c 2 where b 2ac 3.7) a 2 x 2 + bx + c 2 a 2 x 2 + 2acx + c 2 4x x + 9 4x x + 9 ax + c) 2 3.8) 2 2 x x x + 3) 2 If you do not see that a trinomial is a perfect square, that is okay. Just factor it using one of the factoring procedures and you will sucessfully factor the trinomial. It is desirable to spot patterns such as the perfect square because it can greatly speed up factoring. 3.4 Quadratic Formula We should be familiar with the quadratic formula which states if ax 2 + bx + c 0 then x b± b 2 4ac 2a. We let α and β be the two not necessarily distinct solutions to ax 2 + bx + c 0, then ax 2 + bx + c ax α)x β) 3.9) The resulting factors are not necessarily in the most aestetically pleasing form but they are fast to find and it may be easier to polish it afterward than it is to do it the long way. Let s try an example: 14x x 15 Now we will use the quadratic formula: α, β) ) 15), 31 ) ) 15) 2 14) 2 14) , 31 ) , 3 ) ) So 14x x x 5 ) x 3 ) 7 2 x 5 )) 2 7 7x + 5)2x 3) x 3 2 )) 3.11) In this case it was probably easier to factor the trinomial using our original method, but sometimes it can be difficult to find our pair p,q). If finding p and q is difficult, then the quadratic formula gives us an alternative method. 4 Last Updated: June 27, 2012
5 250) term Sometimes it is best to group terms in some slightly more creative ways. These are just a few patterns that may be useful, many more can be found by combining the methods that we have discussed already & 1 Term Grouping 3 & 1 Term Grouping is often used when we have problems with two variables. We ll start with an example: x 2 + 6x + 9 y 2 x 2 + 6x + 9) y 2 x + 3) 2 y 2 x + 3) + y)x + 3) y) x + y + 3)x y + 3) 4.1) In such that example we found a perfect square trinomial inside of the tetranomial, factored it, then noticed that we had a difference of squares and then factored that. Let s look at even a stranger example: x x 4y y + 14 x x 4y y x x y y 9 x x + 25) 4y 2 12y + 9) x + 5) 2 2y 3) 2 x + 5) + 2y 3))x + 5) 2y 3)) x + 2y + 2)x 2y + 8) 4.2) We can generalize this pattern as follows: a 2 x 2 + 2abx α 2 y 2 2αβy + b 2 β 2 ) ax + αy + b + β)ax αy + b β) 4.3) Given the complexity of these pattern, it is often better just to remember the reasoning rather than the generalized formula & 2 Term Grouping 2 & 2 Term Grouping is typically used with expressions in 1 variable with a degree greater than 2. We shall leave the explanation to an example. x 5 + x 4 + x + 1 x 5 + x 4 ) + x + 1) x 4 x + 1) + 1x + 1) x 4 + 1)x + 1) 4.4) 5 Last Updated: June 27, 2012
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