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Mathematics 7 Online
OpenStudy (anonymous):

factor 8x²-16-28x

OpenStudy (anonymous):

4(2x² - 7x - 4)

OpenStudy (amistre64):

put it in "common" form :) 8x^2 -28x -16 = 0 right?

OpenStudy (amistre64):

trop factored the 4, thats good

OpenStudy (amistre64):

we can use quad formula to get the roots; then just reverse it back into a factored form :)

OpenStudy (anonymous):

we have to use AC method

OpenStudy (amistre64):

AC method? air conditioned?

OpenStudy (amistre64):

ohhh....the a*c method; thats my method ;)

OpenStudy (amistre64):

first right down the setup as tho it CAN factor...like this: (x...../a)(x...../a) where a = 2 in this case

OpenStudy (amistre64):

then we can fill in the signs bythe clues given to us in the equation....think we can do it?

OpenStudy (amistre64):

so far: (x /2) (x /2)

OpenStudy (amistre64):

the sign of "C" tells us what to do; the sign of B tells us the sign of the "bigger" number. if the sign of C is (+); we got 2 options: (+)(+) or (-)(-)...do you know why?

OpenStudy (anonymous):

(x+2) (x-2)

OpenStudy (amistre64):

is that an answer? or a response? :)

OpenStudy (anonymous):

it is i have no idea what im doing

OpenStudy (amistre64):

(x /2) (x /2) -16 is (-); we need a (+)(-) to fill in; -28x is (-); the bigger number gets the (-) (x- /2) (x+ /2) lets fill in the blanks now

OpenStudy (amistre64):

A*c = 2(4) = 8; what does 8 factor into? 1,8 8-1=7 <-- lets use these :) 2,4 4-2=2 (x- 8/2) (x+ 1/2) ; reduce till you cant reduce no more... 8/2 = 4/1...stop! 1/2 = 1/2 ....stop! (x-4) (x+1/2) if theres still a fraction; put the bottom # in front.. (x-4) (2x+1)

OpenStudy (amistre64):

its easier to explain in person really :) and with paper...

OpenStudy (anonymous):

very good explanation

OpenStudy (anonymous):

9a^2+ 18a+8

OpenStudy (anonymous):

step by step please

OpenStudy (amistre64):

9 and 8 aint got nothing in common; so we can factor nuthin out to begin with right?

OpenStudy (amistre64):

1) set up as tho it CAN be factored... (a ) (a ) 2) we are going to multiply by that first #, that "9"; so we need to divide it back out in the end like this: (a /9) (a /9) make sense so far?

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

3) look at "signs" to fill in our setup; that last # is a +8, it got there by multiplying 2 numbers together; we know that: (+)(+) = (+) AND (-)(-)=(+) right? so we have 2 options to narrow down 4) that middle #, that +18x; got there by adding or subtracting two numbers together; and we need a (+) result. we know this: (+) + (+) =(+) AND (-) + (-) =(-) So what can we gather from this info? We need 2 signs that are the same to get a (+) last#; AND those 2 signs have to give us a (+) middle #. lets use (+) (+) for that choice like this: (a + /9) (a + /9) We good so far?

OpenStudy (amistre64):

5) lets gather a "pool" of options to choose from now; multiply first and last #s together for this: 9(8) = 72....factor this till you get 2 numbers that add to 18 1,72 ; 72+1 = 73...aint it 2,36 ; ....aint gonna be it either :) 4,18 ... getting closer.... 8,9 ....hmmm, 8+9 = 17....not quite ther 3, 24 ..... 27 is bad 6,12...looks good :) 6+12 = 18 thats it :) we will use 6 and 12 in our setup then like this: (a+ 6/9) (a+ 12/9)

OpenStudy (amistre64):

we reduce those fractions till we can reduce no more... 6/9 = 2/3....stop !! 12/9 = 4/3 ....stop!! (a+ 2/3) (a +4/3) .... whateves got a fraction, but the bottom in front like this: (3a+2) (3a+4) <-- and we are done

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