- #1

Mathsonfire

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Pls hlp I am stuck.Pls give me a clue on this...

Question number 7

Question number 7

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- MHB
- Thread starter Mathsonfire
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In summary, the conversation discusses how to factor a polynomial expression and the best approach to do so. The participants also mention the use of the "@" symbol and how it can trigger the spam detector. One participant suggests a method involving the factorization of (a-b)(b-c)(c-a)(ab+bc+ca), while another participant points out a typo in the last term.

- #1

Mathsonfire

- 11

- 0

Pls hlp I am stuck.Pls give me a clue on this...

Question number 7

Question number 7

Mathematics news on Phys.org

- #2

MarkFL

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MHB

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\(\displaystyle a^2(b^3-c^3)+b^2(c^3-a^3)+c^2(a^3-b^3)\)

I think I would arrange as:

\(\displaystyle c^2(a^3-b^3)-(a^3b^2-a^2b^3)-c^3(a^2-b^2)\)

Factor terms:

\(\displaystyle c^2(a-b)(a^2+ab+b^2)-a^2b^2(a-b)-c^3(a+b)(a-b)\)

Factor out factor common to all 3 terms:

\(\displaystyle (a-b)(c^2(a^2+ab+b^2)-a^2b^2-c^3(a+b))\)

In the second factor, expand and group on like $a$ terms:

\(\displaystyle (a-b)((c^2-b^2)a^2+(bc^2-c^3)a+(b^2c^2-bc^3))\)

\(\displaystyle (a-b)(-(c+b)(b-c)a^2+ac^2(b-c)+bc^2(b-c))\)

\(\displaystyle (a-b)(b-c)(-(c+b)a^2+ac^2+bc^2)\)

\(\displaystyle (a-b)(b-c)((c-a)ac+(a+c)(c-a)b)\)

\(\displaystyle (a-b)(b-c)(c-a)(ac+(a+c)b)\)

\(\displaystyle (a-b)(b-c)(c-a)(ab+bc+ca)\)

- #3

I like Serena

Homework Helper

MHB

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Two of the given answers contain (a-b), so let's try to confirm if (a-b) is a factor or not.

If (a-b) is a factor, then substituting b=a must yield 0... and it does (check it).

Consequently we can already eliminate answer (2).

So we try to extract (a-b). We can extract it from $c^2(a^3-b^3)$ directly, so it makes sense to put that term at the front.

We cannot extract it directly from the first and second terms, so we need to rearrange those so that we can extract (a-b), which is what Mark did.

From cyclic symmetry we can tell that (b-c) and (c-a) will also be factors, or otherwise we can check in the same way by substituting c=b respectively a=c.

Extract those factors as well in the same fashion, which brings us to Mark's final result.

- #4

Mathsonfire

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@I love Serena that was a great method thanks...

@MarkFl thanks again...

@MarkFl thanks again...

- #5

I like Serena

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Let me clean that up.

Edit: Too late. Mark already did. ;)

- #6

MarkFL

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I like Serena said:

Let me clean that up.

Edit: Too late. Mark already did. ;)

A lot of sites now have forum software that allows tagging users with the @ symbol, so I may have to rethink using that as a trigger. :)

- #7

I like Serena

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MarkFL said:A lot of sites now have forum software that allows tagging users with the @ symbol, so I may have to rethink using that as a trigger. :)

Ooh, it would be really cool if typing @ would trigger user name markup somehow (instead of triggering the spam filter). Ah well, that is probably asking for too much.

- #8

MarkFL

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I like Serena said:Ooh, it would be really cool if typing @ would trigger user name markup somehow (instead of triggering the spam filter). Ah well, that is probably asking for too much.

DBTech offers an addon that allows true user tagging using the @ symbol, which I've tested some time ago out of curiosity on my dev site. On XF and IPS sites, typing the @ symbol and one ore more letters will even bring up a popup menu of usernames fitting the pattern typed.

Then the user tagged receives a notification that someone has tagged them, and a link to the post is provided.

- #9

Wilmer

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That "ca" should be "ac" :)MarkFL said:\(\displaystyle (a-b)(b-c)(c-a)(ab+bc+ca)\)

- #10

MarkFL

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Wilmer said:That "ca" should be "ac" :)

I agree, however, I was arranging it the way it is shown in the attached photo. :)

- #11

Wilmer

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A thousand apologies of which you may have one!

Factorisation is the process of breaking down an algebraic expression into smaller parts, known as factors, that when multiplied together result in the original expression.

To factorise an expression with variables, you need to find common factors and use algebraic techniques such as factoring by grouping, difference of squares, or perfect square trinomials.

The steps for factorising an expression include:

- Step 1: Identify any common factors.
- Step 2: Use algebraic techniques to factorise the remaining expression.
- Step 3: Check if the factors can be further simplified.

To factorise an expression with more than three terms, you can use the following steps:

- Step 1: Group the terms in pairs.
- Step 2: Factor each pair separately using the techniques mentioned earlier.
- Step 3: Factor out any common factors between the pairs.
- Step 4: Check if the factors can be further simplified.

Yes, the expression can be further factorised into (a+b+c)(a^2b^2+a^2c^2+b^2c^2-a^2bc-ab^2c-abc^2).

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