Hi How would you factorise this?
4y^2 - 16x^4
Any help, gratefully received.
Maths factorisation?
2007-11-17 2:13 pm
回答 (8)
2007-11-17 2:21 pm
✔ 最佳答案
[01]4y^2-16x^4
=4(y^2-4x^4)
=4{(y)^2-(2x^2)^2}
=4(y+2x^2)(y-2x^2)
2007-11-18 1:31 am
This is the difference of 2 perfect squares.
We know (a^2-b^2) has factors (a+b)(a-b), so
4y^2-16x^4 should behave likewise
4y^2-16x^4=(2y+4x^2)(2y-4x^2)
=2(y+2x^2)* 2(y-2x^2)
=4(y+2x^2)(y-2x^2)
We could have removed the common factor 4 prior to
factoring, to get 4(y^2-4x^4). Perhaps I should have
done so-it's a bit easier.
We know (a^2-b^2) has factors (a+b)(a-b), so
4y^2-16x^4 should behave likewise
4y^2-16x^4=(2y+4x^2)(2y-4x^2)
=2(y+2x^2)* 2(y-2x^2)
=4(y+2x^2)(y-2x^2)
We could have removed the common factor 4 prior to
factoring, to get 4(y^2-4x^4). Perhaps I should have
done so-it's a bit easier.
2007-11-17 10:27 pm
4y^2 - 16x^4
= (2y)^2-((2x)^2)^2
= [2y-(2x)^2][2y+(2x)^2]
= [(sqrt(2y))^2-(2x)^2][2y+(2x)^2]
= [ sqrt(2y)-(2x)][ sqrt(2y)+2x)][2y+(2x)^2]
= 4 [ sqrt(y)-x][ sqrt(y)+x)][y+(x)^2]
= (2y)^2-((2x)^2)^2
= [2y-(2x)^2][2y+(2x)^2]
= [(sqrt(2y))^2-(2x)^2][2y+(2x)^2]
= [ sqrt(2y)-(2x)][ sqrt(2y)+2x)][2y+(2x)^2]
= 4 [ sqrt(y)-x][ sqrt(y)+x)][y+(x)^2]
2007-11-17 10:24 pm
ist simply factor out 4
4 (y^2 -4x^4)
then use the sum and diff of two squares priciples
= 4 (y -2x^2)(y+2x^2)
4 (y^2 -4x^4)
then use the sum and diff of two squares priciples
= 4 (y -2x^2)(y+2x^2)
2007-11-17 10:21 pm
4y^2-16x^4
=(2y)^2-(4x^2)^2
=(2y-4x^2)(2y+4x^2)
=4(y-2x^2)(y+2x^2).ANS.
=(2y)^2-(4x^2)^2
=(2y-4x^2)(2y+4x^2)
=4(y-2x^2)(y+2x^2).ANS.
2007-11-17 10:21 pm
4y^2 - 16x^4
[let 4 be out]
4 (y^2 - 4x^4)
ANS: 4 (y - 2x^2) (y+ 2x^2)
[let 4 be out]
4 (y^2 - 4x^4)
ANS: 4 (y - 2x^2) (y+ 2x^2)
2007-11-17 10:20 pm
(2y - 4x^2)(2y + 4x^2) applying the (a^2 - b^2) = (a+b)(a-b)
2007-11-17 10:18 pm
4y^2 - 16x^4 = 4 (y^2 -4x^4)
= 4 (y -2x^2)(y+2x^2)
= 4 (y -2x^2)(y+2x^2)
收錄日期: 2021-05-03 14:24:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071117061333AAvSbCR