Factoring and/by Grouping problem  TOPIC_SOLVED

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Factoring and/by Grouping problem

Postby dragonwench on Wed Sep 09, 2009 2:52 am

I have a (a few, actually -) problem in a workbook - and I can't seem to factor it out properly:

(**Mathematics 155, Section 1.4 - Miscellaneous Equations)

Directions: Find the real solutions of each equation.

(**Note: in this section, we are specifically working on factoring and grouping along with some radical equations. I know i'm supposed to group/factor somehow but I'm confused on this one problem from this section.)


the problem is written:

(a^2 - 6a)^2 - 2(a^2 -6a) - 35 = 0

Ok, so, First, I factor out a^2 - 6a and group what's remaining b/c I have 4 terms and (a^2 - 6a) is common in both groups...

(a^2 - 6a) (a^2 - 6a) - 2(1) - 35 = 0 (I think this is wrong?)

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(**Afterthought while previewing before hitting send: Should i just distribute everything and THEN factor out everything?)

{So I started over and distributed first and came up with : (a^2 - 6a)^2 - (2a^2 + 12a) - 35 = 0
then factored out : (a^2 - 6a) (a^2 - 6a) - 2 - 2 - 35 = 0

factor out another 'a' and clean it up: a(a-6)(a-6)-39=0
a = 0 , (a-6)(a-6)-39=0
a^2 -12a +36-39 == a^2 - 12a - 3 = 0 is not factorable ... so that didn't work:

and (a-6)^2 = 39 .... and take the square root of both sides.. doesn't work either... (As you can see, I truly have worked this problem many times, in many ways.)

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(a^2 - 6a) (a^2 -6a) - 37 = 0

even if i factor out another a:

a (a-6)(a-6) - 37 = 0

so if i go with this, then i can assume that one solution is going to be a = 0 (so i'll just ignore that 'a' now?)

(a-6)(a-6) - 37 = 0

a^2 -12a +36 - 37 = 0

a^2 - 12a - 1 = 0

so there's no way this can work.... (unless i resort to the quadradic formula, and I don't think that's what he's looking for at this point...)


so if i go back up to my last factoring:

(a-6)(a-6) = 37

a = 43 ?

so my answers are 0, 43?

I still don't think I'm doing this right - and I'm really not sure where I'm going wrong or what I need to correct.

Help would be greatly appreciated.

Thank You,

Stephanie
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  TOPIC_SOLVED

Postby stapel_eliz on Wed Sep 09, 2009 12:47 pm

Please allow me to say "Thank you!!" for showing your work and reasoning so clearly! :clap:

dragonwench wrote:(a^2 - 6a)^2 - 2(a^2 -6a) - 35 = 0

Ok, so, First, I factor out a^2 - 6a and group what's remaining....

You're on the right track in noticing the commonality of the a2 - 6a part. But in this case, since this factor is not common to all three terms, you cannot factor it out front. Instead, note that you have something of the following form:

. . . . .

This factors as:

. . . . .

...to give you the solutions:

. . . . .

Can you see a way of relating the above solution method to your equation? :wink:
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Re: Factoring and/by Grouping problem

Postby dragonwench on Thu Sep 10, 2009 3:11 am

You're welcome - And, Thank You! Thank you for your reply, as well.

We had a review in class today and while he didn't go over this specific problem, he did review similar ones... and it just *clicked* - and I feel like SUCH and idiot... Substitution!!! I needed to substitue a^2 - 6 for a variable... then find the value of the variable then plug it back in and find the real solution. It seems so easy now and I don't know how I missed it. It's as though I just *Forgot* about the option of substitution - I was so focused on factoring and grouping!

Thank you so much for your willingness to help.

I was just directed to this website for the first time, last night. I am sure I will be asking questions as well as trying to help others with questions that I actually understand! :) However, I *am* the type of person... I don't want someone to just give me the answer, I want to Know, Learn, and Understand how to work the problems. It doesn't do me any good to just have answers if I can't work the problems on my own. :) ( That's why I have no problem making it clear that if I'm asking for help, I've already tried everything I can think of and just don't know where I'm going wrong)

I had an exam today and only missed 3 questions. there were 2 questions on finding the distance between two points and two questions on finding the midpoint between two points. I arrived at the correct answer for ONE of the midpoint questions but not the other, and did not arrive at the right answer for either distance problem. I double checked my work on my scratch paper afterwards and I still don't know what I did wrong. I'm wondering if maybe I thought i knew the formulas but don't? Unfortunately, our exams are administered via a computer system and we are not allowed to remove our scratch paper from the Math Lab - so I can't give exact problems...

However, Could you (or someone) tell me if I'm using the wrong formula(s)?

To find the distance between two points on a graph:

d= (Y2-Y1)^2 + (X2-X1)^2 .. all on right side of the = is under the square root sign. (Sorry, I don't know how to SHOW this)

so if two points were.. (I'm pulling this out of the air) (2, 3) and (4, 5)

d= square root of (5-3)^2 + (4-2)^2 ..... d= square root of: 2^2 + 2^2 ..... d= sr of 4+4 so d= Square root of 8 ... SR of 4 * SR of 2 so 2SR2 (if you can decipher what I'm trying to say)


And:

midpoint:

m = ( (Y1+Y2)/2 , (X1+X2)/2 ) so same points as above would be:

m = (2+4)/2, (3+5)/2 ... so m = 6/2 Or 3 and 8/2 Or 4 so the midpoint is : (3, 4)

Am I doing something wrong?

Thank you so much, Again!

I have a few more questions that I'll try to resolve tomorrow on campus as it is VERY late and I've had only 2 hours of sleep in the past 2 days :)

Very Grateful,

Stephanie
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Postby stapel_eliz on Thu Sep 10, 2009 1:10 pm

dragonwench wrote:To find the distance between two points on a graph:

d= (Y2-Y1)^2 + (X2-X1)^2 .. all on right side of the = is under the square root sign. (Sorry, I don't know how to SHOW this)

To learn how to format math as text, try here.

Yes, you are using the correct formula for finding the distance between two points. :thumb:

dragonwench wrote:midpoint:

m = ( (Y1+Y2)/2 , (X1+X2)/2 )....

Am I doing something wrong?

The work you demonstrated here is correct. You may have made a typo on your test...? :oops:
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Re: Factoring and/by Grouping problem

Postby dragonwench on Thu Sep 10, 2009 11:31 pm

Thank You so much. I won't be able to access my exam until sometime next week (after everyone has had an opportunity to take the exam) I'll probably be able to figure out what I missed.

I'm sure I'll be posting again between now and the first week of December :D

Your help is so greatly appreciated.

Thank You, Thank You, Thank you! :)

~S
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