# Binomial Expansions of Polynomial help 😁👍😳

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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.

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.

0

reply

Report

#2

(Original post by

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.

**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.

0

reply

Report

#3

Note that (1 + x - 2x^2) = (1 - x)(1 + 2x), hence (1 + x - 2x^2)^4 = (1 - x)^4.(1 + 2x)^4

0

reply

(Original post by

Is it binomial series or binomial expansion? Could you show the original question?

**ThiagoBrigido**)Is it binomial series or binomial expansion? Could you show the original question?

0

reply

(Original post by

Note that (1 + x - 2x^2) = (1 - x)(1 + 2x), hence (1 + x - 2x^2)^4 = (1 - x)^4.(1 + 2x)^4

**old_engineer**)Note that (1 + x - 2x^2) = (1 - x)(1 + 2x), hence (1 + x - 2x^2)^4 = (1 - x)^4.(1 + 2x)^4

0

reply

Report

#6

(Original post by

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?

**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?

0

reply

(Original post by

Yes, exactly that, and you can ignore all terms in powers of x above x^2.

**old_engineer**)Yes, exactly that, and you can ignore all terms in powers of x above x^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?

0

reply

Report

#8

(Original post by

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?

**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?

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

0

reply

(Original post by

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.

**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.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top