My Math Forum Need help checking/show method.

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 August 2nd, 2017, 05:24 AM #1 Newbie   Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1 Need help checking/show method. Factorise a) [check] 6x^2 - 21xy + 8xz - 28yz = 3x(2x - 7y) + 4z(2x - 6y) b) [method](x - 1)^2 - (y - 2)^2 c) [check] p^2 - y^4 = (p + y^2)(p - y^2) d) [method] 16a^2 - (3b + 4c)^2 e) [check] 3ax + bx - 3ay - by - 9a - 3b = 3ax - 3ay - 9a - 3b + bx - by = 3(ax - ay - 3a - 3b) + b(x - y) Last edited by JoKo; August 2nd, 2017 at 05:45 AM.
 August 2nd, 2017, 05:26 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,453 Thanks: 489 Math Focus: Yet to find out. What have you tried? Let's start with just one question at a time first..
 August 2nd, 2017, 05:55 AM #3 Newbie   Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1 Well, for (d) I expanded the bracket and tried to factorise from there; $\displaystyle 16a^2 - (3b +4c)^2$ $\displaystyle = 16a^2 + 9b^2 + 24bc + 16c^2$ $\displaystyle = 4(4a^2 + 6bc + 4c^2) + 3(3b^2)$
August 2nd, 2017, 07:07 AM   #4
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 Originally Posted by JoKo Well, for (d) I expanded the bracket and tried to factorise from there; $\displaystyle 16a^2 - (3b +4c)^2$ $\displaystyle = 16a^2 + 9b^2 + 24bc + 16c^2$ $\displaystyle = 4(4a^2 + 6bc + 4c^2) + 3(3b^2)$
What happened to your minus sign on the second line .

$- (3b +4c)^2 = -(3b + 4c)(3b + 4c) = -(9 b^2 + 12bc + 12bc + 16c^2) = -1*(9 b^2 + 24bc + 16c^2) = (- 9 b^2 - 24bc - 16c^2)$

 August 2nd, 2017, 07:32 AM #5 Newbie   Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1 Thanks for pointing that out. Is the method correct ?
August 2nd, 2017, 08:19 AM   #6
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 Originally Posted by JoKo Thanks for pointing that out. Is the method correct ?
I think you mis-understand the purpose behind factoring. (Factorizing is just a silly word.)

The purpose is to get an expression into an equivalent expression that is more convenient for some purpose. There may be quite a few equivalent expressions. Which one is "correct" depends on what you want to do.

So your expression for d would have been equivalent and would be therefore A "correct" answer if you had not made that sign error. In other words, your method is valid but not uniquely valid. I suspect that they want a different "correct" answer, namely an answer based on the method of difference of squares.

$16a^2 = (4a)^2. \text{ And obviously } (3b + 4c)^2 \text{ is a square.}$

So

$16a^2 - (3b + 4c)^2 = \{4a - (3b + 4c)\}\{4a + (3b + 4c)\} =$

$(4a - 3b - 4c)(4a + 3b + 4c).$

Last edited by skipjack; August 4th, 2017 at 03:15 AM.

August 2nd, 2017, 08:47 AM   #7
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Quote:
 e) [check] 3ax + bx - 3ay - by - 9a - 3b = 3ax - 3ay - 9a - 3b + bx - by = 3(ax - ay - 3a - 3b) + b(x - y)
$3ax - 3ay - 9a + bx - by - 3b$

$3a(x-y-3) + b(x-y-3)$

$(x-y-3)(3a+b)$

 August 2nd, 2017, 09:04 AM #8 Global Moderator   Joined: Dec 2006 Posts: 18,232 Thanks: 1437 (a) 6x² - 21xy + 8xz - 28yz = 3x(2x - 7y) + 4z(2x - 7y). That leads to (3x + 4z)(2x - 7y). (c) Correct. (b) Use the difference of squares method (as per (c)). (e) 3ax + bx - 3ay - by - 9a - 3b = (3a + b)x - (3a + b)y - 3(3a + b) = (x - y - 3)(3a + b) In hindsight, it can be seen that all the questions have been designed to be easy. It's a good idea to be well aware of the easiest methods and how to see whether they are immediately applicable.

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