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factor 21x^2-49x-9x-21

Mathematics
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idk how i got 7x(3x-7) 3(3x-7) and i know its wrong
is the original problem 21x^2-58x-21 ?
no

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Other answers:

right,so thats 21x^2-40x-21=0 right ?
yeah and then the product of 21 and -21 is -441 and then 40 so the two numbers are -49 and 9
that is correct, see above u have written -49x - 9x instead of what u got now as -49x+9x
oooh i see wow
now whats common from first 2 terms in 21x^2-49x-9x-21 =0?
7x
i got the answer :P -3/7 and 7/3
yup, correct.
im having problems with another equation
which ?
i got to the point 9 + or - sqrt of 1 all over 4
9 from where ?
Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two rots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
yeah b is 9
nopes, b is -4
oh wait nooo i gave you the wrong equation! no wonder
lol :P
still b is -9
but in that equation it says -b so -(-9) is 9
inside i did put put -9^2 = 81
yes, u are correct \(\huge\frac{9 \pm \sqrt{1}}{4}=\frac{10}{4},\frac{8}{4}\)
wait you are just adding and subtracting the 1?
obviously lol never mind i feel stupid
got it now ?
yes, it was a stupid moment :P
ok i did another one and i got 4+-sqrt of 28 all over 2 @hartnn
question ?
its the first one that sent you
thats correct :)
but how do i factor it?
do u want to factor or do u want to use quadratic formula? anyways its not factorable....
so what would be the answer?
i cant simplify it?
all u can do is \(2 \pm \sqrt7\)
\(\sqrt{28}=2\sqrt7\)
how does that equal to that
\(\sqrt{28}=\sqrt{7*4}=\sqrt{7}\sqrt{4}=2\sqrt{7}\)
got it
you are the best!
lol. glad to hear that :) u too are awesome ;)
lol thank you, well i gotta keep doing this stupid hw
best of luck for that.

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