Binomial Expansions of Polynomial help 😁👍😳

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Alexandramartis
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Hello, I have been practising binomial expansions recently and have been using the following practise questions. I believe that I have answered the first two questions correctly, however, the third question is a bit of a stumbling block since it concerns a polynomial not a binomial. However, I have approached a solution and would like advice whether my method to solve this problem is suitable. Thank you to anyone who takes time to reply 👍
For all of the questions below find in their simplest from the first three terms of the expansion of:
Question 1: (1-x)^4
So the binomial theorem states:
(nC0)a^n*b^0+ (nC1)a^n-1*b^1 + (nC2)a^n-2*b^2...
Therefore, where n=4, a =1 and b =-x
(4C0) *1^4*-x^0=1
(4C1)*1^3*-x^1=4*1^3*-x^1=-4x
(4C2)*1^2*-x^2=6*1^2*-x^2=6x^2

Thus, the first three terms of the expansion of (1-x)^4 are: 1 -4x+6x^2...

Question 21+2x)^4
n=4, a= 1 and b =2x
(4C0) *1^4*2x^0=1
(4C1) *1^3*2x^1=8x
(4C2) *1^2*2x^2=24x^2

Thus, the first three terms of the expansion of (1+2x)^4 are: 1+8x+24x^2

Question 3: Hence find the first three terms in ascending powers of x of (1+x-2x^2)^4

Substituting that (x-2x^2)=u
(1+u)^4=1+4u+6u^2+...
Expanding:
1 + 4(x-2x^2)+6(x-2x^2)^2

4(x-2x^2)= 4x-8x^2
6(x-2x^2)^2=6x^2-24x^3+24x^4

Thus, collecting the like terms and simplifying:

1+4x+14x^2-24x^3+24x^4

I realise that the full expansion of (1+x-2x^2)^4 is:
1+4x-2x^2-20x^3+x^4+40x^5-8x^6-32x^7+16x^8

but I cannot show the actual first three terms by only finding the first three terms, since through further expansion the coefficients of e.g. the x^2 or x^3 values change.
Can I please have some advice as to how to properly answer this question, I think the fact that it is a binomial expansion with more than two terms which is confusing me here.
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ThiagoBrigido
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(Original post by Alexandramartis)
Hello, I have been practising binomial expansions recently and have been using the following practise questions. I believe that I have answered the first two questions correctly, however, the third question is a bit of a stumbling block since it concerns a polynomial not a binomial. However, I have approached a solution and would like advice whether my method to solve this problem is suitable. Thank you to anyone who takes time to reply 👍
For all of the questions below find in their simplest from the first three terms of the expansion of:
Question 1: (1-x)^4
So the binomial theorem states:
(nC0)a^n*b^0+ (nC1)a^n-1*b^1 + (nC2)a^n-2*b^2...
Therefore, where n=4, a =1 and b =-x
(4C0) *1^4*-x^0=1
(4C1)*1^3*-x^1=4*1^3*-x^1=-4x
(4C2)*1^2*-x^2=6*1^2*-x^2=6x^2

Thus, the first three terms of the expansion of (1-x)^4 are: 1 -4x+6x^2...

Question 21+2x)^4
n=4, a= 1 and b =2x
(4C0) *1^4*2x^0=1
(4C1) *1^3*2x^1=8x
(4C2) *1^2*2x^2=24x^2

Thus, the first three terms of the expansion of (1+2x)^4 are: 1+8x+24x^2

Question 3: Hence find the first three terms in ascending powers of x of (1+x-2x^2)^4

Substituting that (x-2x^2)=u
(1+u)^4=1+4u+6u^2+...
Expanding:
1 + 4(x-2x^2)+6(x-2x^2)^2

4(x-2x^2)= 4x-8x^2
6(x-2x^2)^2=6x^2-24x^3+24x^4

Thus, collecting the like terms and simplifying:

1+4x+14x^2-24x^3+24x^4

I realise that the full expansion of (1+x-2x^2)^4 is:
1+4x-2x^2-20x^3+x^4+40x^5-8x^6-32x^7+16x^8

but I cannot show the actual first three terms by only finding the first three terms, since through further expansion the coefficients of e.g. the x^2 or x^3 values change.
Can I please have some advice as to how to properly answer this question, I think the fact that it is a binomial expansion with more than two terms which is confusing me here.
Is it binomial series or binomial expansion? Could you show the original question?
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old_engineer
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Note that (1 + x - 2x^2) = (1 - x)(1 + 2x), hence (1 + x - 2x^2)^4 = (1 - x)^4.(1 + 2x)^4
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Alexandramartis
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(Original post by ThiagoBrigido)
Is it binomial series or binomial expansion? Could you show the original question?
I think it is binomial expansion, I have attached the original question to show you anyhow. Thank you for replying 👍
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Alexandramartis
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(Original post by old_engineer)
Note that (1 + x - 2x^2) = (1 - x)(1 + 2x), hence (1 + x - 2x^2)^4 = (1 - x)^4.(1 + 2x)^4
Thnak you for your reply😁Sorry but I am a little unclear, are you suggesting to multiply the values of the individual expansions of (1 - x)^4 and (1 + 2x)^4 e.g. 1 -4x+6x^2 *1+8x+24x^2?
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old_engineer
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(Original post by Alexandramartis)
Thnak you for your reply😁Sorry but I am a little unclear, are you suggesting to multiply the values of the individual expansions of (1 - x)^4 and (1 + 2x)^4 e.g. 1 -4x+6x^2 *1+8x+24x^2?
Yes, exactly that, and you can ignore all terms in powers of x above x^2.
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Alexandramartis
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(Original post by old_engineer)
Yes, exactly that, and you can ignore all terms in powers of x above x^2.
So do you mean (1-4x+6x^2) *(1+8x+24x^2):
1*1=1
1*8x=8x
1*24x^2=24x^2
-4x*1=-4x
-4x*8x=-32x^2
-4x*24x^2=-96x^3
6x^2*1=6x^2
6x^2*8x=48x^3
6x^2*24x^2=144x^4

Collect the like terms and simplify:
1+(8x-4x)+(24x^2+6x^2-32x^2)+(48x^3-96x^3)+144x^4
1+4x-2x^2-48x^3+144x^4

Is this correct?
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old_engineer
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(Original post by Alexandramartis)
So do you mean (1-4x+6x^2) *(1+8x+24x^2):
1*1=1
1*8x=8x
1*24x^2=24x^2
-4x*1=-4x
-4x*8x=-32x^2
-4x*24x^2=-96x^3
6x^2*1=6x^2
6x^2*8x=48x^3
6x^2*24x^2=144x^4

Collect the like terms and simplify:
1+(8x-4x)+(24x^2+6x^2-32x^2)+(48x^3-96x^3)+144x^4
1+4x-2x^2-48x^3+144x^4

Is this correct?
Yes. You needn't have calculated the x^3 and x^4 terms, as the question just asks for the first three terms, i.e. up to the x^2 term. Your original approach was also sound, but you just went astray when collecting the x^2 terms, somehow turning (-8x^2 + 6x^2) into 14x^2.

In questions like this, you should always be on the lookout for ways of re-using results from the early parts of the question later on.
Last edited by old_engineer; 1 year ago
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Alexandramartis
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(Original post by old_engineer)
Yes. You needn't have calculated the x^3 and x^4 terms, as the question just asks for the first three terms, i.e. up to the x^2 term. Your original approach was also sound, but you just went astray when collecting the x^2 terms, somehow turning (-8x^2 + 6x^2) into 14x^2.

In questions like this, you should always be on the lookout for ways of re-using results from the early parts of the question later on.
Excellent, thank you for your help and advice I truly appreciate it immensely! My apologies, I believe that I must have overlooked -8x^2 as being negative, silly me😂
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