Math, asked by sameerthakur673, 1 year ago

Factorise:- x^4 - 4abx^2 - (a^2 - b^2)^2


sameerthakur673: it is a question from akaash
HannanM: bhai recheck kar x2 or x4
HannanM: check this link : https://www.quora.com/How-do-I-factorise-x-2-4abx-a-2-b-2-2

Answers

Answered by IAMHELPINGYOU
0
x^4 - 4abx² - (a² - b²)²
=x^4 - 4abx² - (a+b)²(a-b)²
=x^4 - 4abx² - (a+b)(a+b)(a-b)(a-b)
=x(x³ - 4abx) - (a+b)(a+b)(a-b)(a-b)

If U like it please make it BRAINLIEST

sameerthakur673: but i do not know its process
HannanM: i found the answer i can't write
HannanM: but*
sameerthakur673: you cannot write?
HannanM: no cuz max. no ppl wrote the answer
sameerthakur673: pls. do it in note Book then u can send me picture
HannanM: could you is it x4 or x2
HannanM: check if
sameerthakur673: it is correct . i have right now
sameerthakur673: i have checked it right now
Answered by TPS
5
 {x}^{4} - 4ab {x}^{2} - {( {a}^{2} - {b}^{2} ) }^{2} \\ \\ = {x}^{4} - 4ab {x}^{2} + {(2ab)}^{2} -{(2ab)}^{2} - {( {a}^{2} - {b}^{2} ) }^{2} \\ \\ = {( {x}^{2} - 2ab) }^{2} - 4 {a}^{2} {b}^{2} - ( {a}^{4} + {b}^{4} - 2 {a}^{2} {b}^{2} ) \\ \\ = {( {x}^{2} - 2ab) }^{2} - 4 {a}^{2} {b}^{2} - {a}^{4} - {b}^{4} + 2 {a}^{2} {b}^{2} \\ \\ = {( {x}^{2} - 2ab) }^{2} - {a}^{4} - {b}^{4} - 2 {a}^{2} {b}^{2}

 = {( {x}^{2} - 2ab) }^{2} - ( {a}^{4} + {b}^{4} + 2 {a}^{2} {b}^{2} ) \\ \\ = {( {x}^{2} - 2ab) }^{2} - {( {a}^{2} + {b}^{2})}^{2} \\ \\ = ( {x}^{2} - 2ab + {a}^{2} + {b}^{2} )( {x}^{2} - 2ab - {a}^{2} - {b}^{2}) \\ \\ = [ {x}^{2}+ {(a-b)}^{2}][ {x}^{2}- {(a+b)}^{2}]\\ \\ = [{x}^{2}+ {(a-b)}^{2}][(x+a+b)(x-a-b)]\\ \\ = [{x}^{2}+ {(a-b)}^{2}](x+a+b)(x-a-b)

sameerthakur673: so I got this question
HannanM: prinitng error can be also possible
sameerthakur673: is not printed it is a PDF
HannanM: Typing error
HannanM: are you having you pt2 exam
HannanM: your*
sameerthakur673: pt2 exam?
HannanM: periodic test 2
HannanM: i'm my maths exam tom. i was just go through your question
TPS: I think my answer is correct. still, i will ask someone to verify
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