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OpenStudy (anonymous):
Please hepl!
OpenStudy (ajprincess):
which of the options give u the expression 16x^4 – b^4 + 10b^2 – 25 when distributed
OpenStudy (anonymous):
distributive property?
OpenStudy (ajprincess):
a(b+c)=a*b+a*c-(Distributive property)
OpenStudy (ajprincess):
i dnt thnk that's a good idea.
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OpenStudy (anonymous):
Hmm ok. I'm confused :(
OpenStudy (ajprincess):
@hartnn
OpenStudy (ajprincess):
really sorry for confusing u
hartnn (hartnn):
16x^4 – b^4 + 10b^2 – 25
= 16x^4 - (b^4 -10 b^2 +25)
can you first try to factor (b^4 -10 b^2 +25) ??
hint : its a perfect square ?
OpenStudy (anonymous):
No no I am sorry for bothering you
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OpenStudy (anonymous):
Ok I will try
OpenStudy (anonymous):
Is it b^6-15
OpenStudy (anonymous):
@hartnn
hartnn (hartnn):
how you got that ?
hartnn (hartnn):
ok,for factorizing the standard quadratic equation,\( ax^2+bx+c=0\), we need 2 numbers whose sum is 'b' and product = ac
for your Question ,\( (b^4 -10 b^2 +25) = ( (b^2)^2 -10 (b^2) +25)\)
can you find 2 numbers whose sum = -10 and product = 25 ?
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OpenStudy (anonymous):
-5 and -5
hartnn (hartnn):
yes, so can you factorize that now by writing -10b^2 = -5b^2-5b^2
??
OpenStudy (anonymous):
Yes
hartnn (hartnn):
what u get after factoring ?
OpenStudy (anonymous):
I don't know:(
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hartnn (hartnn):
(b^4−5b^2-5b^2+25)
now factor out b^2 from first 2 terms and -5 from last 2 terms, what u get ?
OpenStudy (anonymous):
(b^2-5)^2
hartnn (hartnn):
yes, thats correct :)
so now you have
\(16x^4-(b^2-5)^2 = (4x^2)-(b^2-5)^2\)
use the difference of squares formula \(a^2-b^2=(a+b)(a-b)\) to factor that
OpenStudy (anonymous):
16x^4 + b^4 - 50
OpenStudy (anonymous):
I think it's not correct :(
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hartnn (hartnn):
use 4x^2 as 'a' and (b^2-5) as 'b'
OpenStudy (anonymous):
so 4x^2-(b^2-5)
hartnn (hartnn):
that is just a-b
what about a+b ?
OpenStudy (anonymous):
4x^2-(b^2-5) = 4x^2+(b^2+5)
OpenStudy (anonymous):
Is it correct?
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hartnn (hartnn):
what that = in between ?
its
(4x^2-(b^2-5))( 4x^2+(b^2-5))
= .....?
OpenStudy (anonymous):
oh sorry :(
OpenStudy (anonymous):
so it is (4x^2-(b^2-5))( 4x^2+(b^2-5))
hartnn (hartnn):
thats what i typed, one step of simplification and you'll get your answer.
OpenStudy (anonymous):
(4x^2 – b^2 – 5)(4x^2 + b^2 – 5)
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