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kaylala
Group Title
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.
 one year ago
 one year ago
kaylala Group Title
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.
 one year ago
 one year ago

This Question is Closed

kaylala Group TitleBest ResponseYou've already chosen the best response.0
See the document posted above for a CLEARER VIEW of the GIVEN question
 one year ago

chmvijay Group TitleBest ResponseYou've already chosen the best response.0
can you expand it by multiplication
 one year ago

shkrina Group TitleBest ResponseYou've already chosen the best response.1
r there powers
 one year ago

kaylala Group TitleBest ResponseYou've already chosen the best response.0
i dont know how? @chmvijay
 one year ago

shkrina Group TitleBest ResponseYou've already chosen the best response.1
Well first add the like terms .. in both groups ...
 one year ago

shkrina Group TitleBest ResponseYou've already chosen the best response.1
IS this ur question \[((x^{2}+3y)+y ^{2})((x ^{2}+3y)y ^{2}\])
 one year ago

shkrina Group TitleBest ResponseYou've already chosen the best response.1
hellooo.........
 one year ago

kaylala Group TitleBest ResponseYou've already chosen the best response.0
yes, it is @shkrina
 one year ago

shkrina Group TitleBest 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
 one year ago

kaylala Group TitleBest ResponseYou've already chosen the best response.0
so, what's the product or the final answer?
 one year ago

whpalmer4 Group TitleBest 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 \(a = (x^2+3y) , b = y^2\)
 one year ago

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

shkrina Group TitleBest ResponseYou've already chosen the best response.1
@kaylala did u finish this sum....
 one year ago

kaylala Group TitleBest ResponseYou've already chosen the best response.0
not sure @shkrina
 one year ago

shkrina Group TitleBest ResponseYou've already chosen the best response.1
well .dw:1372577408636:dw
 one year ago

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

shkrina Group TitleBest ResponseYou've already chosen the best response.1
simplify u will get the ans ///
 one year ago

kaylala Group TitleBest ResponseYou've already chosen the best response.0
@shkrina but this is my answer: x4+6x2−y4+9y2
 one year ago

shkrina Group TitleBest 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 .....
 one year ago

whpalmer4 Group TitleBest 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.
 one year ago

shkrina Group TitleBest 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 .........
 one year ago
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