SOLUTION: 16y^2-8y+1

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Question 337623: 16y^2-8y+1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you want to factor this


Looking at 16y%5E2-8y%2B1 we can see that the first term is 16y%5E2 and the last term is 1 where the coefficients are 16 and 1 respectively.

Now multiply the first coefficient 16 and the last coefficient 1 to get 16. Now what two numbers multiply to 16 and add to the middle coefficient -8? Let's list all of the factors of 16:



Factors of 16:
1,2,4,8

-1,-2,-4,-8 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 16
1*16
2*8
4*4
(-1)*(-16)
(-2)*(-8)
(-4)*(-4)

note: remember two negative numbers multiplied together make a positive number


Now which of these pairs add to -8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -8

First NumberSecond NumberSum
1161+16=17
282+8=10
444+4=8
-1-16-1+(-16)=-17
-2-8-2+(-8)=-10
-4-4-4+(-4)=-8



From this list we can see that -4 and -4 add up to -8 and multiply to 16


Now looking at the expression 16y%5E2-8y%2B1, replace -8y with -4y%2B-4y (notice -4y%2B-4y adds up to -8y. So it is equivalent to -8y)

16y%5E2%2Bhighlight%28-4y%2B-4y%29%2B1


Now let's factor 16y%5E2-4y-4y%2B1 by grouping:


%2816y%5E2-4y%29%2B%28-4y%2B1%29 Group like terms


4y%284y-1%29-1%284y-1%29 Factor out the GCF of 4y out of the first group. Factor out the GCF of -1 out of the second group


%284y-1%29%284y-1%29 Since we have a common term of 4y-1, we can combine like terms

So 16y%5E2-4y-4y%2B1 factors to %284y-1%29%284y-1%29


So this also means that 16y%5E2-8y%2B1 factors to %284y-1%29%284y-1%29 (since 16y%5E2-8y%2B1 is equivalent to 16y%5E2-4y-4y%2B1)


note: %284y-1%29%284y-1%29 is equivalent to %284y-1%29%5E2 since the term 4y-1 occurs twice. So 16y%5E2-8y%2B1 also factors to %284y-1%29%5E2



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Answer:
So 16y%5E2-8y%2B1 factors to %284y-1%29%5E2


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website.

Jim