Factor the following expression completely: -6x^3 – 20x^2 + 16x
GCF?
4
Did you tried anything ?
not quite, look at the x's plus, what is -6/4 ? not an integer.
Let's also factor out the negative sign and an _?
\( -6 x^3 - 20 x^2 + 16 x =-2 x (4+x) (-2+3 x) \)
The GFC is 2
Hey fool I was trying to figure it out
u forgot the x sinz
-2x is the GFC try to confirm FFM's answer... maybe he's wrong.
lol, yeah I could be wrong.
human errors do occur..
Yeah, provided I am human though ...
-2x(4+x)(-2+3x) = -6x^3 - 20x^2 + 16x
yeah he was right?
well that's one way to confirm FFM's answer I was hoping you'd try to pull out a -2x though, for practice on factoring.
yeah I am but when you have 6 different people telling you how to do it it is hard
-6x^3 - 20x^2 + 16x =-2x(3x^2)-2x(10x)-2x(-8)=-2x(3x^2+10x-8) ca you finish?
can*
-6x^3 - 20x^2 + 16x =-2x(3x^2)-2x(10x)-2x(-8)=-2x(3x^2+10x-8)= -6x^3 - 20x^2 + 16x right?
-2*3 is -6 -2*10= 20 -2*-8 = 16 then you add the different powers together
I meant the exponents
Yes but not quite what I asked you to do... I want to see if you can get FFM's answer on your own. so we have -2x(3x^2+10x-8) can you factor the (3x^2+10x-8) part? that will either confirm or deny the answer FFM gave once and for all.
typo... what is 3x^2+10x-8 factored?
okay lets see (2x+x)(5-4) right?
or is it backwards
should it be (2x+5)(x-4)
no... \[(2x+x)(5-4)=3x(1)=3x\neq3x^2+10x-8\]you need one x term iun each set of parentheses ccloser, but you wont get the 3x term from above\[(2x+5)(x-4)=2x^2+x-20\]so keep trying.
one thing you know is that because x^2 has a 3 next to it, the factorization will be\[(3x\pm?)(x\pm?)\]because that is the only way to get the three to show up...
okay (3x + 4)(x + 2) does this look right?
multiply it out and see
\[(3x+4)(x+2)=3x^2+2x+4x+8=3x^2+6x+8\] in order to get the -20 to show we need at least to have\[(3x\pm4)(x\pm5)\]or the other way\[(3x\pm5)(x\pm4)\]that way we make sure we have the 3 at the beginning and the -20 at the end...
okay but to factor 3x^2 + 10x - 8 it Factors to: (3x - 2)(x + 4) = 0
A good strategy is to look at the first and last terms. That tell you a lot about how it can be factored. Let's try yours\[(3x-2)(x+4)=3x^2-2x+12x-8=3x^2+10x-8\]so it looks like that works! (disregard what I said about the -20, I forgot what we were trying to factor :/ ) now does that give the same answer as FFM ?
yes it does
Well I guess it was all for nothing then, might as well just take his word blindly from now on... Yeah right!!!
I think so he has it backwards though
right, but it is ok he took the negative of each term in the parentheses and (-1)(-1)=1 so they are the same
he has it to where it is (4+x)(-2+3x) so no it doesn't add up to the -20
for the x it will Like I said I got confused and forgot what we were trying to factor what I meant is that the produce of the two constants should be -8 like in the original, which he has.
okay so he was right then?
product* lol yes he was right multiply both you and his answer out and you get the same
your* cannot type today
okay so do you think I did okay on the factoring part then?
I thin you need practice. If you want that get a deck of 3x5 cards and write out various products like (x+2)(x+4)=x^2+6x+8 (2x-5)(x+7)=2x^2+2x-35 (3x-1)(2x+2)=6x^2+4x-2 (just write the right-hand side, the result of the multiplication) and write them out onto the card. Then shuffle them and see if you can run the process backwards: x^2+6x+8=??? 2x^2+2x+35=??? 6x^2+4x-2=??? when you can do that then you have factoring down!
think*
Alright that sounds like it should be fun
right? enjoy :D
you too and thank you again
Join our real-time social learning platform and learn together with your friends!