Select one of the factors of 5x2 + 7x + 2.
(5x - 2)
(x + 2)
(5x + 1)
None of the above
@hartnn I want to try and do this one on my own but I was wondering if I get something wrong then you could guide me back?
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hartnn (hartnn):
sure,
start!
OpenStudy (anonymous):
okay :) lol
OpenStudy (anonymous):
so first you would have to find what makes +2 and 14?
hartnn (hartnn):
\(5x^2 +7x+2\)
2 numbers with SUM = +7
and PRODUCT = 5*2 = 10
?
OpenStudy (anonymous):
oh picked the wrong number
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hartnn (hartnn):
for \(ax^2+bx+c\)
it is
sum = +b
product = ac
OpenStudy (anonymous):
okay lol
hartnn (hartnn):
you can remember as `sum`= middle number
product = product of `extreme` numbers
hartnn (hartnn):
** `middle`
OpenStudy (anonymous):
okay i wrote that down :)
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OpenStudy (anonymous):
so you would have to find what makes 10 and 7
hartnn (hartnn):
yes
sum as 7
and product as 10
OpenStudy (anonymous):
so \[5+2=7 and 5\times2=10\]
hartnn (hartnn):
correct :)
so how will we split 7x as ?
OpenStudy (anonymous):
so \[(5+2)(5\times2)?\]
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hartnn (hartnn):
we split the middle term right ?
we got +5 and +2
so we split 7x as
\(\Large 7x = +5x+2x\)
hartnn (hartnn):
\(\Large 5x^2 +5x+2x+2\)
OpenStudy (anonymous):
oh okay
OpenStudy (anonymous):
so okay now im stuck
hartnn (hartnn):
as always
consider first 2 terms
\(\Large 5x^2+5x\)
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hartnn (hartnn):
what can be factored out ?
OpenStudy (anonymous):
x(5x+5)?
OpenStudy (anonymous):
wait i think i did it wrong
hartnn (hartnn):
x can be factored out, yes
but 5 can be factored out too, right ?
OpenStudy (anonymous):
oh yes i was a little confused on that lol
so
\[5x \left( x+1 \right)\]
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hartnn (hartnn):
correct
now
2x+2
OpenStudy (anonymous):
2(x+1)
OpenStudy (anonymous):
so it would be............
OpenStudy (anonymous):
\[(5x+2)(x+1)\]
hartnn (hartnn):
excellent! :D
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OpenStudy (anonymous):
but what would be the answer tho? im looking at them but idk
OpenStudy (anonymous):
oh wait it would be d
hartnn (hartnn):
its not from the first 3 options
so its None :)
hartnn (hartnn):
yes,
you got d correct ;)
OpenStudy (anonymous):
thank you for helping me!!!!
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