Hi everyone. I was wondering if anyone could help me answer this question. The mark scheme just shows an answer but I don’t know how to get there.

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

Original post by #Daisy

Hi everyone. I was wondering if anyone could help me answer this question. The mark scheme just shows an answer but I don’t know how to get there.

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

A - you have a difference of two squares (hint)

Original post by #Daisy

Hi everyone. I was wondering if anyone could help me answer this question. The mark scheme just shows an answer but I don’t know how to get there.

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

Assuming youve done A, then B is similar but youve got a bit of partial factorisation to do first. You have three numbers which are repeated, so partially factorise and evaluate those two terms and it should be reasonalby clear.

Original post by #Daisy

Show how binomial expansion can be used to solve the following without a calculator:

A: (268^2)-(232^2)

B: (65.1*29.2+35.9*65.1-26.4*91.7+26.4*65.3)/(18.3^2 -18.3*5.4)

Any help would be amazing.

Hi Daisy,

For A, you can consider writing it as (250+18)^2 - (250-18)^2 and use binomial expansion of (250+x)^2 and then use x = 18 and -18 and simplify

Original post by mqb2766

A - you have a difference of two squares (hint)

No, although you are right, you won't be accurate in this situation since it was asked to make use of binomial expansion.

For part B) It can be written as (65.1^2-26.4^2)/(18.3*12.9) and use binomial expansion to simplify further

Original post by PiyushRaghav

No, although you are right, you won't be accurate in this situation since it was asked to make use of binomial expansion.

Of course it will be accurate - it's an exact method.

Original post by PiyushRaghav

For part B) It can be written as <redacted> and use binomial expansion to simplify further

If (as mqb suggested) you use difference of two squares, you subsequently simply need to spot common factors.

(edited 5 months ago)

Original post by DFranklin

Of course it will be accurate - it's an exact method.

Tbf, its probably what the question asked for, though its less obvious than just dots. For the OP the question is asking about inverse of babylonian multiplication

https://en.wikipedia.org/wiki/Babylonian_mathematics#Arithmetic

so evaulate 4ab rather than (a+b)^2 - (a-b)^2. Though obviously dots/simple factorisation is easier to spot and gives the "same" 4ab term.

(edited 5 months ago)

Original post by mqb2766

Tbf, its what the question asked for, though its less obvious than just dots. For the OP the question is asking about inverse of babylonian multiplication

https://math.widulski.net/worksheets/BabylonianMultiplication.html

so evaulate 4ab rather than (a+b)^2 - (a-b)^2. Though obv dots/simple factorisation is easier to simply spot and gives the "same" 4ab term.

https://math.widulski.net/worksheets/BabylonianMultiplication.html

so evaulate 4ab rather than (a+b)^2 - (a-b)^2. Though obv dots/simple factorisation is easier to simply spot and gives the "same" 4ab term.

Partially because I see no point in making life difficult, I would claim here that since technically, (a-b) and (a+b) are binomial terms, (a-b)(a+b) = a^2-b^2 is a binomial expansion.

Original post by DFranklin

Partially because I see no point in making life difficult, I would claim here that since technically, (a-b) and (a+b) are binomial terms, (a-b)(a+b) = a^2-b^2 is a binomial expansion.

Agreed and for me its the easiest/obvious thing to do.

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