# Math Help - Binomial Theorem expanded

1. ## Binomial Theorem expanded

ok i just want someone to double check my work please. for some reason i think i have the wrong answer.

(use the binomial theorem to expand and simplify this expression)

(x-y)^8

ok so i used pascal's triangle ...the pascal triangle in my book stops at the 7th row so i got this for my 8th row 1,8,28,56,70,56,28,8,1

so this is what i got ...Please check my work.

1x^8-8x^7y+28x^6y^2-56x^5y^3+70x^4y^4-56x^3y^5+28x^2y^6-8xy^7+1y^8

ok i know this looks confusing but if you copy and paste it into online calculator it should put everything into place. Did i do this right???

2. If I was marking I's say "no", because it says "use the binomial theorem", not "read the appropriate row of Pascal's Triangle from your textbook".

I reckon what you ought to do would be to use the identity

$\binom n m = \frac {n!} {(n-m)! m!}$

where $n = 8$ and $m$ is each of 0 to 8 in turn. The arithmetic is not hard, and the numbers should work out the same as what you got.

3. Originally Posted by Matt Westwood
If I was marking I's say "no", because it says "use the binomial theorem", not "read the appropriate row of Pascal's Triangle from your textbook".

I reckon what you ought to do would be to use the identity

$\binom n m = \frac {n!} {(n-m)! m!}$

where $n = 8$ and $m$ is each of 0 to 8 in turn. The arithmetic is not hard, and the numbers should work out the same as what you got.
i dont understand what your trying to tell me. its asking me to expand (x-y)^8

what you gave me is no where in my book? your technique probably works too but im just no familiar with it sorry.

4. Originally Posted by flexus
ok i just want someone to double check my work please. for some reason i think i have the wrong answer.

(use the binomial theorem to expand and simplify this expression)

(x-y)^8

ok so i used pascal's triangle ...the pascal triangle in my book stops at the 7th row so i got this for my 8th row 1,8,28,56,70,56,28,8,1

so this is what i got ...Please check my work.

$1x^8-8x^7y+28x^6y^2-56x^5y^3+70x^4y^4-56x^3y^5+28x^2y^6-8xy^7+1y^8$

ok i know this looks confusing but if you copy and paste it into online calculator it should put everything into place. Did i do this right???
HI

you are correct .

Instead of using the pascal triangle to look for the coefficient , you can actually use the identity which matt has suggested .

HI

you are correct .

Instead of using the pascal triangle to look for the coefficient , you can actually use the identity which matt has suggested .
could you show me how to do that? if its not to much trouble that is? thank you.

6. im sorry i still don't get it so did i do it right or wrong?? is my answer correct?

7. Show you how to do what? Multiply and divide? Matt told you that the coefficient of $x^ny^{8-n}$ is $\frac{8!}{n!(8- n)!}$. Calculate that for n= 0 to 8 (Actually, you only need to do it for n= 0 t0 4, after that you get duplicates). Do you know that $n!= n(n-1)(n-2)\cdot\cdot\cdot (3)(2)(1)$? So that The coefficient for $x^2y^6$ is $\left(\begin{array}{cc}8 \\ 2\end{array}\right)= \frac{8!}{2! 6!}= \frac{40320}{(2)(720)}$. You can simplfy that a lot by observing that $\frac{8!}{2! 6!}= \frac{8(7)(6)(5)(4)(3)(2)(1)}{(2(1))(6(5)(4)(3)(2) (1)}$ and canceling a lot- you eventually get 4(7)= 28, just as you have!

8. im sorry but no one has told me if i did it right or if i got the right answer???

9. Originally Posted by flexus
im sorry but no one has told me if i did it right or if i got the right answer???
The answer is correct. But note that when a question gives an instruction to solve it in a particular way, your solution must show that this instruction has been followed ....