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happykiddoBest ResponseYou've already chosen the best response.0
i know the answer is (x8) (2x3).........but how exactly would i get that?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
2x^219x+24 First multiply the coefficient of x^2 and the constant. That is, the coefficient of x^2 here is 2 nd constant is 24. 2*24=48 Then find two factors of 48 such that their sum is 19 and product is 48. That is if the factors are a and b a+b=19 a*b=48
 one year ago

happykiddoBest ResponseYou've already chosen the best response.0
yeah but 24 and 2 dont add up to 19 or any of the other multiples i can think of for 48
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
I think the expression u have given shoul be 2x^211x+48 and nt 2x^219x+48. plzz chck
 one year ago

happykiddoBest ResponseYou've already chosen the best response.0
no its still 2x^219x+24
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
ya u r right. Am sorry. Let me list out the factors of 48 1,2,3,4,6,8,12,16,24,48,1,2,3,4,6,8,12,16,24,48. Nw chck if u can find those numbers
 one year ago

happykiddoBest ResponseYou've already chosen the best response.0
so the answer should be (2x3) (x16)..........not what the book gives me which is (x8) (2x3)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
2x^219x+24 =2x^216x3x+24 =2x(x8)3(x8)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
After finding the numbers u have to split only the mid term using thm. Gettng t? @happykiddo
 one year ago

happykiddoBest ResponseYou've already chosen the best response.0
oh i see where your going with this THANK YOU :D
 one year ago

happykiddoBest ResponseYou've already chosen the best response.0
this really helped.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
Glad to hear that nd welcome:)
 one year ago

happykiddoBest ResponseYou've already chosen the best response.0
yay i just got two more problems right. i dont think i'll be needing more help : )
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
Congrats for gettng ur problems rght:)
 one year ago
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