View Full Version : Help im a dumbass ;O need math help...
x²-7x+12
10x²-3xy-18y²
can u guys show me how to "Factory Completely (Remove the greatest common factor as needed)" ? ;[ i know im stupid
is the complete factorization of x²-7x+12
(x-3)(x-4)? can i take it further? am i completely wrong?
x²-7x+12
10x²-3xy-18y²
I actually enjoy math so Ill help you out. The first is (x -4)(x -3). Second is (5x - 6y)(2x + 3y).
The first type is real easy, they follow the (x +- #1)(x +- #2) format. To get the 7x part you add #1 and #2, to get the 12 you multiply #1 and #2. Make sure to keep the sign (+ or -) of each # into consideration.
I could be horribly wrong though I havent taken a math class in like 6 years. :p Why are you doing math in the summer time btw? Summer school?
(x-3)(x-4)
(5x+6y)(2x-3y)
@Jest...Your solution yields the wrong sign for 3xy
(x-3)(x-4)
(5x+6y)(2x-3y)Ah damn he owned me I messed up the signs. :p
yea i havent taken a math class in 2 years and highest i had in highschool was Algebra 1, now im taking college algebra and trig which... after 2 years of math and that math only being a dumbass's highschool math... yea ;P
nah im in a tech school, all year round
TheGreatMilenko
03-08-04, 21:14
x²-7x+12
10x²-3xy-18y²
can u guys show me how to "Factory Completely (Remove the greatest common factor as needed)" ? ;[ i know im stupid
whats Factory Completely ?
ok u guys im really bad w/ math can you guys please explain it reallly... reallly... dumbasslike >.<
jest you said you add the coefficient of x² and 12 but i thought ur supposed to * it.
i thought it was something like
A² - b + c
A*C = # and 2 factors of that number * together = b
and the highest factor ( in this case 4 goes in the first thing ie:
(x - 4) then lowest factor goes in the second things ( x - 3 )
but how did you get 10x²-3xy-18y²'s answer? can you please go step by step >.< im slow as hell
edit: u werent even talkin about this u were talkin about foil, yea i know foil to check it but i just dont how to get to that point >.<
atm im doing 4x²-31x-8
ok so 4*8 = 32, the difference( since the first sign is minus ) of factors of 32 that equal are 32 and 1. so
(2x-32)(2x+1) ???
edit: nope doesnt foil out
(2x-1)(2x+32)
4x²+64x-3x-32
4x²+61x-32
doesnt foil out right either ;[ what am i doing wrong? should there be no 2 in front of the x's? no there has to be to get it to foil correctly ;[... ;.; what am i doing wrong
atm im doing 4x²-31x-8
ok so 4*8 = 32, the difference( since the first sign is minus ) of factors of 32 that equal are 32 and 1. so
(2x-32)(2x+1) ???
edit: nope doesnt foil outAnswer is (4x + 1)(x - 8). Tbh I just do it in my head, which most teachers require you to show work. Lemme see if I can remember an equation or something.
(4x+1)(x-8)
Tbh I just do it in my head..Me too. Can't remember the actual procedure. That was a long time ago. :D
heh now im royally screwed... 3x²+20z-8 / 4z²+21z+20
do i factor the top and bottom then ??
ok woots, im starting to figure this stuff out >.<
the top is (3z-2)(z+4) and bottom is (4z + 5) (z+4) w00ts ;] but took me freakin 10 mins
now gotta 7x² / 4 * 28 / x^3
guessing that s 49/x^3
ok thx u guys for helpin me out ive finally figured it all out and do it all relativly fast now. my comp prof helped me alot too ;P she did it fast as hell every time i asked her so i was like @_@.... and she finally slow'd down and had to remember how to do it out instead of her head like u guys ;P
This is the kinda stuff i was learning in school about 2 years ago. I don't think there's actually any definitive fool proof way of doin em, just gotta use trial and error, and a bit of lateral thinking.
Wait until you see implicit intergration of logarithms
thats when the fun really begins
Jeez this thread has brought back memories ............ I'll have to did out my old Maths text books and burn em :p
ax²+bx+c = 0
When there is no first coefficient (a=1), it's fastest and most practiced to be done manually. This is a common process:
x²-7x+12 (no first coefficient)
1) We want two numbers that add to the second coefficient (-7) and that multiply to the third coefficient (+12).
2) Start with the third coefficient, +12. It has factors of 1:12, 2:6, 3:4...
3) The first two factors cannot provide us a -7, but the third can, -3 and -4 add to -7.
4) So your solution is (x-3)(x-4)
When there is a first coefficient (a!=1), you use the quadratic equation:
_________
-b ± √ b² - 4ac
x = --------------- to find your roots (x-root1)(x-root2)
2a
For 3z²-10z-8:
________________
10 ± √ -10² - 4*3*(-8) 10 ± 14
z = ---------------------- -> z = ------- -> z = 4, -2/3 (roots)
2*3 6
With your roots: (z-4) and (z+2/3) or (3z+2) after multiplying it by 3.
Test: (z-4)(3z+2) = 3z²+2z-12z-8 = 3z²-10z-8
For 4z²+21z+20:
_____________
-21 ± √ 21² - 4*4*20 -21 ± 11
z = -------------------- -> z = -------- -> z = -5/4, -4 (roots)
2*4 8
With your roots: (z+4) and (z+5/4) or (4z+5) after multiplying it by 4.
Test: (z+4)(4z+5) = 4z²+5z+16z+20 = 4z²+21z+20
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