Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Look for the special product of:
[( x2 + 3y) + y2] [(x2 + 3y) – y2]
NOTE:
~ See the clearer given in the comment area
~ Group some terms to find the applicable special product.
~Follow the chain rule of special product.
~ The Final answer should be free from all symbols of grouping.
 9 months ago
 9 months ago
Look for the special product of: [( x2 + 3y) + y2] [(x2 + 3y) – y2] NOTE: ~ See the clearer given in the comment area ~ Group some terms to find the applicable special product. ~Follow the chain rule of special product. ~ The Final answer should be free from all symbols of grouping.
 9 months ago
 9 months ago

This Question is Closed

kaylalaBest ResponseYou've already chosen the best response.0
See the document posted above for a CLEARER VIEW of the GIVEN question
 9 months ago

chmvijayBest ResponseYou've already chosen the best response.0
can you expand it by multiplication
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
i dont know how? @chmvijay
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
Well first add the like terms .. in both groups ...
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
IS this ur question \[((x^{2}+3y)+y ^{2})((x ^{2}+3y)y ^{2}\])
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
well if so multiply every term of group ((x2+3y)+y2) with all elements of group ((x2+3y)+y2) .... done
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
so, what's the product or the final answer?
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.2
My guess is that you are supposed to think of this as \[(a+b)(ab) = a^2b^2\]with \(a = (x^2+3y) , b = y^2\)
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
ok @whpalmer4 is the answer equal to: \[x ^{4}+6x ^{2}y ^{4}+9y ^{2}\]
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
@kaylala did u finish this sum....
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
well .dw:1372577408636:dw
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
u will get \[x ^{4}+3x ^{2}yx ^{2}y ^{2}+3yx ^{2}+9y ^{2}3y3+x ^{2}y ^{2}+3y ^{3}y ^{4}\]
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
simplify u will get the ans ///
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
@shkrina but this is my answer: x4+6x2−y4+9y2
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
ss rt ans ....the above eq is the eq u get after multiplying ....u can cancle 3y^3and x^2y^2/.....and add ..3x^2y ... u will get the same ans ....hope clear .....
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.2
@shkrina you disregarded the instructions for this problem. Whether or not there's a different way to solve a problem, if the instructor asks for it to be done a specific way, that's the way it should be done for the purposes of learning the material. @kaylala as I said \[(a+b)(ab) = a^2b^2\] with \(a=(x^2+3y),b=y^2\) Write out \[(a+b)(ab) = a^2b^2\]That is the "special product" Now replace \(a\) on the right hand side with \((x^2+3y)\): \[a^2b^2 = (x^2+3y)^2 b^2\]Now replace \(b\) on the right hand side with \(y^2\) \[a^2b^2=(x^2+3y)^2(y^2)^2\] That was the "chain rule" of special product Now we have to expand everything to get rid of the grouping symbols. Let's do the easy one first: \[(x^2+3y)^2(y^2)^2 = (x^2+3y)^2y^4\]because \[(u^n)^m = u^{n*m}\]Now we expand the first product term by repeatedly applying the distributive property \[(x^2+3y)^2y^4 = (x^2+3y)(x^2+3y)y^4 = x^2(x^2+3y)+3y(x^2+3y)  y^4\]\[=x^2*x^2+x^2*3y + 3y*x^2+3y*3y  y^4 \]\[=x^4+3x^2y+3x^2y+9y^2y^4\]Now collect like terms \[=x^4+6x^2y+9y^2y^4\] So your answer was very close, but you somehow lost the \(y\) portion of the \(6x^2y\) term. You should revisit your work and try to figure out why you made that mistake.
 9 months ago

shkrinaBest ResponseYou've already chosen the best response.1
@kaylala ,, @whpalmer4 well sorry ....i did not pay much attention to the variable ...... this time i will look through........ once again sorry ,,,,,, @kaylala in u r ans 6x^2y ,,,,not 6x^2....u missed y...... hope its clear .........
 9 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.