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factor out the 'w;' and work the quad
do I have to have the w on the outside or can I put it in form (w-4)(w+2)
the 'w' is a part of the factoring; it needs to be a part of the soultion or you dont get the same answer. That would be like saying; if I dont pay you for the first 3 hours you work, do you still get a full paycheck :)
or numbers for example: 4(3)(2) = 24 ; can we omit the 4? ___(3)(2) = 6; cant do it ;)_
ok well my work shows there are no letters on the outside of the parenthesis on any of the questions that are the same. So that is why I'm trying to figure out why urs does and theres dont
are the problems you are looking at start with a 'w^3' ? 'w' being a dummy variable and any letter will work.
the highest exponent in a problem is a good indicator of how many factors it will have
x^2 can only have at most 2 factors x*x x^3 can only have at most 3 factors: x*x*x
this is exactly what they give me Factor the trinomial w^3-2w^2-8w there showing me to pair of factors and sum of factors to get the answer
first step; factor out all common elements: the 'w' w ( w^2 -2x -8 ) step 2: factor, if possible, the inside of the (...) w (w-4) (w+2) step 3: factor inside the (..) if possible. no other factors step 4: and.
lol.... and means end :)
what two numbers equal 2, and they have to be common factor of both 2 and 8
-4(2) = -8 ... check -4+2 = -2 ... check
(x+b)(x+c) = xx + xc +bx +bc x^2 +(b+c)x + b*c is how we form a quadratic
to factor it, we need to find 'b' and 'c' that make the middle and last numbers
x^2 + (-4+2)x + -4(2) x^2 -2x -8 ; you just gotta be able to see how its made... and that takes practice :)