## Factoring and/by Grouping problem

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
dragonwench
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### Factoring and/by Grouping problem

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

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

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:

. . . . .$Y^2\, -\, 2Y\, -\, 35\, =\, 0$

This factors as:

. . . . .$(Y\, -\, 7)(Y\, +\, 5)\, =\, 0$

...to give you the solutions:

. . . . .$Y\, =\, 0\, \mbox{or}\, Y\, =\, -5$

Can you see a way of relating the above solution method to your equation?

dragonwench
Posts: 6
Joined: Wed Sep 09, 2009 2:03 am
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### Re: Factoring and/by Grouping problem

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

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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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.

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...?

dragonwench
Posts: 6
Joined: Wed Sep 09, 2009 2:03 am
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### Re: Factoring and/by Grouping problem

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

Your help is so greatly appreciated.

Thank You, Thank You, Thank you!

~S