18x^2 + 37xy - 20y^2?

Question

could someone please explain how u factor that?? i know the answer bc i have an answer key, but i have no idea how 2 get it

Answer 1

18x^2 + 37xy - 20y^2

to factor this you have to do some trial factoring
(9x - 4y)(2x + 5y)
You haveto find a number that will multiply to 18x^2 , so first find out what multiples will give you 18, which would be 6 times 3=18 or 9 times 2=18 or 18 times 1=18, you haveto plug each of these into the equation, secondly, to find the second number it hasto multiply out to -20 and add up to your midterm of 37xy, so to get -20 I used 5y and -4y, sometimes you haveto switch the order around that's why it's called trial factoring. You haveto use the foil method in which first you multiply the first terms which are the x's then you multiply the outside, which is 9x times 5y and then you do the inside terms of -4y times 2x, the last term will be -4y times 5y which is - 20y^2.
(9x - 4y) (2x + 5y)
18x^2 + 45xy - 8xy -20y^2
combine like terms and you get the answer of
18x^2 + 37xy - 20y^2

Answer 2

Think of x/y=t. Then your equation becomes 18 t^2+ 37 t - 20=0
You can solve it like any second degree equation and then you can write this as 18( t - a)(t - b) where a and b are the roots you found. Finally you plug in t = x/y xhich gives you, after mulitiplication by y^2,
18 (x-ay)(x-by). This is your factorization.

Answer 3

I think Gialino has the best answer. You can also assume y=1
The question becomes 18x^2+37x-20. You can factor it.
(9x-4)(2x+5)
Now add y back to the answer.
The true anwer is
(9x-4y)(2x+5y)

Answer 4

18x^2 + 37xy - 20y^2

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-37 ± sqrt(37^2 - 4(18)(-20)))/(2(18))
x = (-37 ± sqrt(1369 + 1440))/36
x = (-37 ± sqrt(2809))/36
x = (-37 ± 53)/36
x = (16/36) or (-90/36)
x = (4/9) or (-5/2)

ANS : (9x - 4y)(2x + 5y)

Answer 5

5 seriose!

Answer 6

if you think about this
(x+y)(x+y)=x^2 +2yx+y^2
Thus
(9x-4y)(2x+5y)