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You should use quadratic formula for this.
but I have to use factoring.
factoring means to undo multiplication
we have to put this into a form that can be unproducted lol
a product is the result of multiplying; to undo multiplication means to factor it out again
3*2 produces 6 6 factors into 3 and 2
Well I sort of factored but I have questions..
4y^2 - 11y - 3 = 4y^2 + y - 12y - 3 = y(4y + 1) - 3(4y + 1) = (4y + 1)(y - 3)
there are a few algorithms for factoring quadratics; pratu did factor by grouping there :)
how about this:
But the point is: how does it help in solving the equation?
4y^2 - 12y + y - 3=0 4y (y-3) + (y-3) = 0 (y-3) (4y+1) = 0
same results :)
but... I don't get why it's "(y-3)(4y + 1) =0
like from 4y (y-3)+(y-3) to (y-3)(4y + 1) =0
a(b+c) = ab + ac ab + ac = a(b+c)
your fatoring out a (y-3) in this case
ohhhh. factoring a y-3?
what do you mean?
grouping allows you to "see" the common factor of (y-3) right?
then factor it out ....
(4y + 1)(y - 3) = 0 |dw:1328452809130:dw|
ohhh.. I think I'm getting it, ugh. Why is math so difficult for me >_<
you have 2 terms: 4y (y-3) and 1(y-3) the GCF is (y-3)
ohh, it is.
4(2) + 1(2) is the same as 2(4+1) right?
but why is it (y-3)(4y+1) now? o.o
well, in this case it aint a 2, its (y-3)
= (-y + 4)/(y - 3)
Does it help in any way?
4y = (-y + 4)/(y - 3)
yes it does
thank you guys!