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Mathematics 7 Online
OpenStudy (anonymous):

Factor 16x^4 – b^4 + 10b^2 – 25 completely. A. (4x^2 + b^2)(2x + b)(2x – b) + 5(2b^2 – 5) B. (4x^2 – b^2 – 5)(4x^2 + b^2 – 5) C. (2x – b^2 + 5)(2x + b^2 – 5)(4x^2 + b^2 – 5) D. (4x^2 – b^2 + 5)(4x^2 + b^2 – 5)

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.

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

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 ?

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:(

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 :(

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?

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)

OpenStudy (anonymous):

Is it correct?

hartnn (hartnn):

yes

OpenStudy (anonymous):

Thank you very much :)

hartnn (hartnn):

welcome ^_^

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