check my work pleaseee for factoring
rationals* not factoring
i think it's just your missing parenthesis....
no, i just did it i get the same answer
-2 , -5/7 is the correct answer to \(\large 7x^2+19x+10=0 \)... so what's the problem?
that's not what im getting i just put it there by using the quadratic function on my calc
how are u solving that quadratic?
factoring?
yes
well since 7 is prime you know that has to be the first term in one of the binomials... (7x + )(x + ) = \(\large 7x^2+19x + 10 \) now just play around with the factors of 10 that will give you 19x in the middle term of the trinomial...
14 and 5 are the factors
but what do you do with the coeff ?
14 is not a factor of 10
can you list all the factors of 10 for me?
it's -2
-2 will give you zero
is that an answer to one of my questions?
which one?
what what are you asking me, im confused
you need to figure out what goes in the ???: (7x + ???)(x + ??) = \(\large 7x^2+19x+10 \) and i said you need to play around with the factors of 10.... my question is... what are the factors of 10?
1,2,5,10
ok... good... can we agree that 1 and 10 will not work here?
nope
no, we cannot agree... or no, it won't work?
no 1 and 10 wont work
ok... it's the 2 and 5 we're concerned with... which gives a correct expansion (first or second one?): (7x + 2)(x + 5) or (7x + 5)(x + 2) of \(\large 7x^2+19x+10 \) ???
(7x + 5)(x + 2)
ok.... there's your factorization....
thankyouu so very much !
yw...:)
since you already knew your answer (from your calculator), you could actually work backwards to get your factorization: since you know -2 was a solution, then (x + 2) has to be a factor... do that with the other solution you'll get (7x + 5)....
ohhh ! yeah i didnt think of that, that would have been much easier
i didn't wanna say anything because i'm a masochist!
...:)
hahaha !
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