Maths : How to factorise it ?
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Need help to factorise this question
a^3+3(a^2)*b+2(a^2)*c+3ab^2+4abc+ac^2+b^3+2(b^2)*c+b*c^2
Any advice / help/ hint is greatly appreciated.
Thanks.
Managed to factorise this question a^2+4b^2+c^2 - 4ab - 4bc + 2ac = a^2 +(- 2b)^2 +c^2 +2(a)(- 2b) +2(- 2b)(c) + 2(a)(c) = (a - 2b + c)^2 but how to apply the concept to the above question or are there any method to factorise this type question easily other than using taking out common factor, grouping, difference in squares, trial and error method, factorising a cubic expression by finding the first factor - these methods cannot seem to be able to factorise the above question.
Wolfram can factorise the above question but the program does not show the steps or method used to factorise the above question.
The multinomial theorem only shows the expansion but it does not show how the expanded terms can be factorised back to its original form.
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I observe that there is a^3+3(a^2)b+3ab^2+b^3, which is actually (a+b)^3.
By factorising c from the remaining expression, I get
2a^2 + 4ab + 2b^2 + bc + acI notice that the first 3 terms give me 2(a+b)^2, while the last 2 terms give me c(a+b).
To summarise, the factorisation is (a+b)^3 + 2c(a+b)^2 + c^2(a+b)
If I wanna try to factorise further, I can factorise (a+b) out first. Then I will get
(a+b) [ (a+b)^2 + 2c(a+b) + c^2 ]If u stare hard at the 2nd big bracket, u will realise that it is in the form of x^2 + 2xy + y^2, which is actually (x+y)^2. So the final factorisation is
(a+b)(a+b+c)^2Tell me if I made any mistakes here. I didn't use any special technique other than being familiar with the expansions of (x+y)^2 and (x+y)^3. Could u tell me where u got this question from? Thanks!
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Deleted
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Factorise completely x^2 - 4x - 12y^2
Expand (3a - b + c^3)^3 (a - 3b + c+ 5d - 6e + 7f + 8g - 9h^15)^12
Factorise completely
162x^9 - 243x^8y^3 + 432x^7y - 1080x^7z - 648x^6y^4 + 1620x^6y^3z
+ 432x^5y^2 - 2160x^5yz + 2700x^5z^2 - 648x^4y^5 + 3240x^4y^4z
- 4050x^4y^3z^2 + 192x^3y^3 - 1440x^3y^2z + 3600x^3yz^2
- 3000x^3z^3 - 288x^2y^6 + 2160x^2y^5z -5400x^2y^4z^2
+ 4500x^2y^3z^3 + 32xy^4 - 320xy^3z +1200xy^2z^2 - 2000xyz^3
+ 1250xz^4 - 48y^7 + 480y^6z - 1800y^5z^2 +3000y^4z^3 - 1875y^3z^4
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Originally posted by Seowlah:
Factorise completely x^2 - 4x - 12y^2
Expand (3a - b + c^3)^3 (a - 3b + c+ 5d - 6e + 7f + 8g - 9h^15)^12
Factorise completely
162x^9 - 243x^8y^3 + 432x^7y - 1080x^7z - 648x^6y^4 + 1620x^6y^3z
+ 432x^5y^2 - 2160x^5yz + 2700x^5z^2 - 648x^4y^5 + 3240x^4y^4z
- 4050x^4y^3z^2 + 192x^3y^3 - 1440x^3y^2z + 3600x^3yz^2
- 3000x^3z^3 - 288x^2y^6 + 2160x^2y^5z -5400x^2y^4z^2
+ 4500x^2y^3z^3 + 32xy^4 - 320xy^3z +1200xy^2z^2 - 2000xyz^3
+ 1250xz^4 - 48y^7 + 480y^6z - 1800y^5z^2 +3000y^4z^3 - 1875y^3z^4
Dont hao lian.....
Nobody but Wolfram supercomputer will solve this kind of sums.
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Originally posted by 238:
Dont hao lian.....
Nobody but Wolfram supercomputer will solve this kind of sums.
Question 1 can be solved easily without the help of Wolfram.
Question 2 is not solved by writing the first bracket 3 times and the second bracket 12 times and then multiply each of the inside terms even without the help of Wolfram.
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Originally posted by Snoopyies:
Need help to factorise this question
a^3+3(a^2)*b+2(a^2)*c+3ab^2+4abc+ac^2+b^3+2(b^2)*c+b*c^2
Any advice / help/ hint is greatly appreciated.
Thanks.
Managed to factorise this question a^2+4b^2+c^2 - 4ab - 4bc + 2ac = a^2 +(- 2b)^2 +c^2 +2(a)(- 2b) +2(- 2b)(c) + 2(a)(c) = (a - 2b + c)^2 but how to apply the concept to the above question or are there any method to factorise this type question easily other than using taking out common factor, grouping, difference in squares, trial and error method, factorising a cubic expression by finding the first factor - these methods cannot seem to be able to factorise the above question.
Wolfram can factorise the above question but the program does not show the steps or method used to factorise the above question.
The multinomial theorem only shows the expansion but it does not show how the expanded terms can be factorised back to its original form.
Heres my standard solution to the question
http://www.freefilehosting.net/eqn_1
You dont need to rely on "do it yourself" methods such as pattern spotting or grouping to solve problem. Simply follow the steps, and any such problems can be solved without fail.
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Originally posted by 238:
Heres my standard solution to the question
http://www.freefilehosting.net/eqn_1
You dont need to rely on "do it yourself" methods such as pattern spotting or grouping to solve problem. Simply follow the steps, and any such problems can be solved without fail.
link is dead:
here:
https://skydrive.live.com/redir?resid=721A3D70B40A20EF!167&authkey=!AACPiok7YLXh-_k
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Originally posted by 238:
Again, the link is dead.
Dont bluff people leh.
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Originally posted by 238:
Ha, ha.
238 says "you dont need to rely on "do it yourself" methods such as pattern spotting or grouping to solve problem. Simply follow the steps, and any such problems can be solved without fail" and
give 2 dead links.
Talk big only. Bluff people only.
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Wow, Seohlah, I think you can "下山" and start your maths class liao : http://www.curica.com
>>Snoopyies and 238 , I thought that local university have their free maths clinic for their student, why don't you drop by for your questions ?
1) NUS Maths Clinic : http://ww1.math.nus.edu.sg/clinic.htm
2) NTU Maths Clinic : http://www.spms.ntu.edu.sg/mas/Document/News/Math Clinic announcement_Notice Board0910.pdf
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Originally posted by M the name:
Wow, Seohlah, I think you can "下山" and start your maths class liao : http://www.curica.com
>>238 , I thought that local university have their free maths clinic for their student, why don't you drop by for your questions ?
1) NUS Maths Clinic : http://ww1.math.nus.edu.sg/clinic.htm
2) NTU Maths Clinic : http://www.spms.ntu.edu.sg/mas/Document/News/Math Clinic announcement_Notice Board0910.pdf
Hi M the name,
Forumer 238 is not asking the question. But he or she claims that anyone who follows his or her steps given in the TWO DEAD LINKS can factorise the polynomial of degree n with multiple variables.
Forumer 238 talks big and bluff people only lah.
Forumer 238 cannot even factorise the simple first question x^2 - 4x - 12y^2,
YET he or she claims to be able to factorise polynomial of degree n with multiple variables.
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To illustrate the different approaches of factorisation
Question
Factorise completely x^3 + 3x^2 - 4
Approach 1
x^3 + 3x^2 - 4
= x^3 + 3x^2 - 1 - 3
= x^3 - 1 + 3x^2 -3
= (x -1)(x^2 +x+1) +3(x^2 - 1)
= (x -1)(x^2 +x+1 +3(x - 1)(x + 1)
= (x - 1)[x^2 + x + 1 + 3(x +1)]
= (x - 1)(x^2 + 4x + 4)
=(x - 1)(x + 2)^2
Approach 2
x^3 + 3x^2 - 4
= x^3 - x^2 + 4x^2 - 4
= x^2(x -1) + 4(x^2 -1)
= x^2(x -1) + 4(x -1)(x + 1)
= (x - 1)[x^2 + 4(x+1)]
= (x - 1)(x^2 + 4x + 4)
=(x - 1)(x + 2)^2
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Appraoch 3
x^3 + 3x^2 - 4
= 4x^3 - 3x^3 + 3x^2 - 4
= (4x^3 - 4) - 3x^3 + 3x^2
= 4(x^3 - 1) - 3x^2 (x - 1)
= 4(x^3 - 1^3) - 3x^2 (x - 1)
= 4(x - 1)(x^2 + x + 1) - 3x^2 (x - 1)
= (x - 1) [4(x^2 + x + 1) - 3x^2]
= (x - 1)(x^2 + 4x + 4)
=(x - 1)(x + 2)^2
Approach 4
x^3 + 3x^2 - 4
= x^4 - x^4 + x^3 + 3x^2 - 4
= x^4 + 3x^2 - 4 - x^4 + x^3
= (x^2 + 4)(x^2 - 1) - x^3(x - 1)
= (x^2 + 4)(x + 1)( x - 1) - x^3(x - 1)
= (x - 1)[ (x^2 + 4)(x + 1) - x^3]
= (x - 1)(x^2 + 4x + 4)
=(x - 1)(x + 2)^2
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Appraoch 5
x^3 + 3x^2 - 4
= x^3 + 3x^2 - 4x + 4x - 4
= x^3 + 3x^2 - 4x + 4(x - 1)
= x(x^2 + 3x - 4) + 4(x - 1)
= x(x - 1)(x + 4) + 4(x - 1)
= (x - 1)[x(x + 4) + 4]
= (x - 1)(x^2 + 4x + 4)
=(x - 1)(x + 2)^2
Approach 6
Take out a common factor first
x^3 + 3x^2 - 4
= (x - 1) (x^2 + 4x + 4)
=(x - 1)(x + 2)^2