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please show me how to factor::: 2x^2-19x+24

Mathematics
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i know the answer is (x-8) (2x-3).........but how exactly would i get that?
2x^2-19x+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
yeah but 24 and 2 dont add up to -19 or any of the other multiples i can think of for 48

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Other answers:

I think the expression u have given shoul be 2x^2-11x+48 and nt 2x^2-19x+48. plzz chck
no its still 2x^2-19x+24
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
hw abt -3 and -16
so the answer should be (2x-3) (x-16)..........not what the book gives me which is (x-8) (2x-3)
2x^2-19x+24 =2x^2-16x-3x+24 =2x(x-8)-3(x-8)
After finding the numbers u have to split only the mid term using thm. Gettng t? @happykiddo
oh i see where your going with this THANK YOU :D
this really helped.
Glad to hear that nd welcome:)
yay i just got two more problems right. i dont think i'll be needing more help : )
Congrats for gettng ur problems rght:)

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