4(x/y+y/x)^2 -23(x/y+y/x)-6
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Factorization of the expression ?
let (x/y + y/x) = z
4 z² - 23 z - 6
= 4 z² + z - 24 z - 6
= z (4 z + 1) - 6 (4 z + 1)
= (z - 6) (4 z + 1)
= (x/y + y/x - 6) ( 4x/y + 4y/x + 1)
let (x/y + y/x) = z
4 z² - 23 z - 6
= 4 z² + z - 24 z - 6
= z (4 z + 1) - 6 (4 z + 1)
= (z - 6) (4 z + 1)
= (x/y + y/x - 6) ( 4x/y + 4y/x + 1)
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Answer:
Here is your answer
Let x/y+y/x be a then
4a²-23a-6
4a²+a- 24a - 6
By regrouping
a(4a+1) -6(4a+1)
(4a+1) (a- 6)
4(x/y+y/x)+1 [(x/y+y/x) - 6]
Final answer
(4x/y+4y/x+1) (x/y+y/x - 6)
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