Factor completely

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Gr8fu13

Junior Member
Joined
Feb 13, 2011
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The problem to factor completely is:
64r^2 + 1 - 16r

I figured I had to get in the format of ax^2 + bx + c
So I have it as:
64r^2 - 16r + 1

I then tried to find factors for 64^2
(1 * 16)
(2 * 32)
(4 * 16)

I tried to fill these in:
(1r )(16r )
(2r )(32r )
(4r )(16r )

I could not find numbers that equal 1 when multiplied and -16 when added. Could these answers be a fraction maybe?
 
Gr8fu13 said:
The problem to factor completely is:
64r^2 + 1 - 16r

I figured I had to get in the format of ax^2 + bx + c
So I have it as:
64r^2 - 16r + 1

I then tried to find factors for 64^2
(1 * 16)
(2 * 32)
(4 * 16)

I tried to fill these in:
(1r )(16r )
(2r )(32r )
(4r )(16r )

I could not find numbers that equal 1 when multiplied and -16 when added. Could these answers be a fraction maybe?
Click to expand...

(-8r) * (-8r) = 64r[sup:dpx4e57n]2[/sup:dpx4e57n] * 1

(-8r) + (-8r) = -16r
 
OR: 64r^2 - 16r + 1 = 0
Use the quadratic formula
If you can't factor "easily", ALWAYS, repeat ALWAYS, use the quadratic formula :idea:
 
Thanks for all your help. I keep forgettting about the quadratic formula. I sometimes have a hard time using that too. For this particular problem I ended up with a discriminent of 0. So since the discriminent of 0 that means there is only one real solution, right? That would be (-8r)(-8r)? When the discriminent is zero I would just divide -b from the 2(b) under the quadratic formula, is that correct? I had -16/128 (since the discriminent was 0) and that is how I came up with my answer of -8. I have never seen an answer like that in any of our text assignments. It's usually (x+3)(x-3) just as an example. That is what I think of when I think of factoring.
 
Gr8fu13 said:
Thanks for all your help. I keep forgettting about the quadratic formula. I sometimes have a hard time using that too. For this particular problem I ended up with a discriminent of 0. So since the discriminent of 0 that means there is only one real solution, right? That would be (-8r)(-8r)? When the discriminent is zero I would just divide -b from the 2(b) under the quadratic formula, is that correct? I had -16/128 (since the discriminent was 0) and that is how I came up with my answer of -8. I have never seen an answer like that in any of our text assignments. It's usually (x+3)(x-3) just as an example. That is what I think of when I think of factoring.
Click to expand...

You are correct. The discriminant is equal to zero - giving you "repeated roots".

64r[sup:1uwfo9y6]2[/sup:1uwfo9y6] - 16r + 1 = (8r - 1)[sup:1uwfo9y6]2[/sup:1uwfo9y6]

Once you have learnt the quadratic formula - other methods are used very rarely.
 
So which would be the correct form to write the answer in?
(-8r)(-8r)
0r
(8r-1)^2
or
(8r-1)(8r-1)

The instructions were: Factor completely
 
Gr8fu13 said:
So which would be the correct form to write the answer in?
(-8r)(-8r)
0r
(8r-1)^2
or
(8r-1)(8r-1)

The instructions were: Factor completely
Click to expand...

You can check to see which of those give the original expression when multiplied out.

You started with 64r[sup:353h6pnl]2[/sup:353h6pnl] - 16r + 1. If you don't GET that when you multiply your "factors" together, then you've not factored correctly.

And, I hope you realize that (8r - 1)[sup:353h6pnl]2[/sup:353h6pnl] means exactly the same thing as (8r - 1)(8r - 1). Whether you choose to write that with an exponent or not may be a matter of some concern to your instructor (ESPECIALLY if you need to enter your answer into some dumb computer program)....it would not matter to me.
 
Yes, I do have to type it into a computer:( I am just going to go with (8r-1)^2 and see what happens. Thanks!
 
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