Simple Algebra Factoring guide
How to do Algebra Factoring
Algebra factoring, factors, factoring, factorization...all the same thing.
The word factorization comes from the word factors. Once you understand what factor means you will soon realize that factorization just applies the theory of it to algebra type problems.
Factor means: Well, by example, the factors of a number such as 12, would be 1, 2, 3, 4, 6 and 12. So, these numbers are the factors of 12.
How did we work that out? Its all the numbers that divide into 12 exactly. So, the morale of the story or sums rather, is...it helps to know your multiplication tables really well to be able to do algebra factoring.
Now, where algebra factoring is concerned, we use the word factorize or factorization. For example, we might say:
Factorize the expression 2a+4
Looking at 2a+4, we want to find what is common or is a factor of both 2a and 4.
Factors of 2 would be 1 and 2 .
Factors of 4 would be 1, 2 and 4.
Factors of a would be a.
Which is common in the 2 groups (disregarding the number 1 for the moment)? The answer would be 2 of course.
If there are letters, we also have to take that into consideration too. In this case there is an 'a'. It is only with the number 2 and not with the 4. Therefore 'a' is not common in both of them.
We open up a set of brackets and put the common number on the outside, in this case 2. This number multiplies the stuff that goes inside the bracket. So that when or if we do multiply it out we return back to the original answer.
Double check this 2(a+2), to see if we return back to 2a+4
Yes it works! This algebra factoring is done!
Factors of 21, 1, 3, 7 .
Factors of 14, 1, 2, 7 .
Factors of b, would be b.
Factors of a, would be a.
So the common factor in 21b+14a would be 7. Put the number 7 on the outside of a bracket. Within the bracket is the multiple of the number, so that when we multiply 7 with a number we return back to the multiples 21 and 14. In this case it would be 3 and 2.
Lets double check this 7(3b+2a), to see if we return back to 21b+14a
Looks like it works fantastically!
Factors of 15 would be 1, 3, 5 , 15
Factors of 20 are 1, 2, 4, 5, 10, 20
Factors of xy are x, y , xy
Factors of y are y
The common number and letter? 5 and y of course. So we put these two on the outside of the bracket 5y( ). To decide what goes in the inside, when you multiply 5y with something it should return back to the original answer. So 5y×3x=15xy and similarly 5y×4=20y.
Factors of 28 are 1, 2 , 4 , 7, 14, 28.
Factors of 16 are 1, 2 , 4 , 8, 16.
Factors of t² would be t² and t
Factors of ts are t , s and ts.
So, what's common out of these numbers and letters? 2, 4 and t of course. But how do we choose out of 2 and 4? Well, going back to the above, where I said to disregard the number 1...lets come to that now.
If we were to choose the number 1 as the common factor, then 1 would be stuck on the outside...oh, 1 multiplied by any thing is that any thing.
Example 1 would be 1(2a+4)=(2a+4)
Example 2 would be 1(21b+14a)=(21b+14a)
and so on and so forth. So what's the point in making 1 the main factor, its not doing anything to the expression apart from adding brackets.
Lets go the the original question, of which one to choose out of 2 and 4.
Lets do both. Lets choose 2 first as the common factor, and t of course. So, putting 2t on the outside of the bracket.
Then we get 2t(14t+8s). What do you notice about the inside part of the bracket? Yes, it can factorize again, there is another common factor within 14 and 8.
So this is the point...it helps to choose the biggest factor on the outset. Continuing with factoring 14 and 8, the common factor would be 2 again. So, taking this out of the equation and combining it with the already present 2t, we then get 4t on the outside and (7t+4s) on the inside.
Final solution would be 4t(7t+4s).
In this question we had to factorize twice. It could have been avoided if we just chose the biggest factor which was 4. Then we would have only factorized it once.
Factors of 9 are 1, 3, , 9.
Factors of 18 are 1, 2, 3, 6, 9 , 18.
Factors of w² would be w² and w .
Factors of a are a.
The common factor in all of them would be 9, a and w.
Join them together and put them outside a bracket 9aw( ). To find the inside numbers or letters. What do we need to multiply 9aw by to get back to the answer?
Why do we need to do factorization? When we do higher level mathematics, we use factorization to manipulate and eventually solve equations. When we take factors out of the equation it simplifies the thing and hence makes it easier to solve. More on this later!
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