dividing polynomials

zee-f · 2011-08-16 07:19:35

just checking to see if i did this right

the original problem says Divide 15x^7 45x^5 / 3x^4

i got 5x^3-15x

i divided 15x^7 by 3x^4 i got 5x^3

multiplied 5x^3 by 3x^4
got 15x^7
i put down the zero and drag the -45x^5
that goes in 3x^4
15x^1 times 3x^4 is ganna give me 45x^5
subtract am left with 0

bobbym · 2011-08-16 08:12:33

That is correct!

zee-f · 2011-08-17 01:08:36

thanx i have this second one am stuck on to

Divide (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)
so i cross out the x-4 they all divide to give me one

left with (3x-2)-(6-5x)/(8x-1)

is this step right ?

Bob · 2011-08-17 01:54:14

Stay on-line while I have a look.

Ok, I've looked. There's a little thing you have missed. The bottom factor isn't (x-4); it is (4-x)

So you should think of it as -(x-4).

When you cancel there'll be a minus on the bottom which will change the signs of the bottom terms.

Bob

Last edited by Bob (2011-08-17 01:57:26)

zee-f · 2011-08-17 02:21:12

so i would get

(3x-2)-(6-5x)\-(x-4)(8x-1)

ill divide that know

zee-f · 2011-08-17 03:02:43

??

bobbym · 2011-08-17 03:04:51

Hi zee-f;

(3x-2)-(6-5x)\-(x-4)(8x-1)

You did that thing again. You can not replace / with \

zee-f · 2011-08-17 03:07:26

i was typing fast i missed that //// lol

(3x-2)-(6-5x)/-(x-4)(8x-1)

bobbym · 2011-08-17 03:13:13

Hi;

Have you copied the problem correctly? This is what I get.

zee-f · 2011-08-17 03:14:30

i didnt start dividing yet i just wanted to know if that step was correct

zee-f · 2011-08-17 03:18:09

so from (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)
to (3x-2)-(6 -5x)/-(x-4)(8x-1)=(8 x-8)/(4-x) (8 x-1)

bobbym · 2011-08-17 03:23:58

Hi zee-f;

1) (3x 2)(x - 4) (x - 4)(6 5x) / (4 x)(8x 1)

2) (3x-2)-(6 -5x)/-(x-4)(8x-1)

1 to 2 is not correct.

Bob · 2011-08-17 03:45:42

hi zee-f

Sorry, I went off.

Back to post #5. The (x-4) should be cancelled away so that

there'll be a minus on the bottom

That is a minus and the other bracket only.

Bob

reconsideryouranswer · 2011-08-17 04:12:58

zee-f wrote:

just checking to see if i did this right
the original problem says Divide (15x^7 45x^5)/(3x^4)

zee-f wrote:

Divide [(3x 2)(x - 4) (x - 4)(6 5x)] /[(4 x)(8x 1)]
so i cross out the x-4 they all divide to give me one
left with [(3x-2)-(6-5x)]/(8x-1)

zee-f wrote:

so from [(3x 2)(x - 4) (x - 4)(6 5x)] /[(4 x)(8x 1)]
to [(3x-2)-(6 -5x)]/[-(x-4)(8x-1)] *to* (8x-8)/[(4-x)(8 x-1)]

zee-f, you must use grouping symbols, as in the above where I
have amended in these quote boxes (parentheses, brackets, or
braces) because of the Order of Operations.

zee-f · 2011-08-17 08:11:16

[(3*x-2)*(x-4)-(x-4)]*[(6-5*x)/(4-x)]*(8*x-1) = -(37*x^2-39*x-2)

zee-f · 2011-08-17 08:15:44

ok i willl thanx

so
[(3x 2)(x - 4) (x - 4)(6 5x)] /[(4 x)(8x 1)]

[(3x 2)(x - 4) (x - 4)(6 5x)]
when i factor it am ganna get
8(x-4)(x-1)

and [(4 x)(8x 1)] factored is

-(x-4)(8x-1)

bobbym · 2011-08-17 08:33:28

Hi;

Is this the problem you are working on?

zee-f · 2011-08-17 10:05:20

yup thats the problem

zee-f · 2011-08-17 10:37:47

hi:

ok so i tried to multiply the top part first using the foil method
this is what i got

[(3x-2)(x-4)] - [(x-4)(6-5x)]

f: 3x*x
o: 3x*4
i: 2*x
l: -2*-4

(3x^2 - 12x - 2x - 8 )

[(x-4)(6-5x)]

f: x*6
o:x*5x
i:-4*6
l:-4*-5x

(6x-5x^2-24+20x)

[(3x^2 - 12x - 2x - 8 )] - [(6x-5x^2-24+20x)]

[(-3x^2+-12x+-2x+-8)] - [(-5x^2+-20x+-6x+-24)]

2x^2+8x-8x+16

the bottom is ganna be from (4-x)(8x-1)
to

[(33x-4-8x^2)]

Last edited by zee-f (2011-08-17 10:58:25)

zee-f · 2011-08-17 11:13:51

am getting really confused so i watched this video
http://www.youtube.com/watch?v=l6_ghhd7kwQ&feature=relmfu

really interesting but still confused because this polynomial its longer and just reaaaaaaaalllllllllllllly confusssing can somebody plzzzzzzz help thanx

bobbym · 2011-08-17 13:44:48

Hi zee-f;

Hold it! No need to do that.

In math you have to be tricky. Trickier than a politician even.

Now do you remember the trick bob bundy showed you in post #4 ?

zee-f · 2011-08-17 22:34:31

hi bobbym

yup i remember so i would get [(3x-2)(6-5x)] / [-(x-4)(8x-1)] ?

bobbym · 2011-08-17 22:38:13

Hi;

You did not cancel on the bottom!

zee-f · 2011-08-17 22:40:37

so [(3x-2)(6-5x)] / [-(x-4)] ?

bobbym · 2011-08-17 22:41:52

Hi;

Nope, you cancelled the wrong term.

Math Is Fun Forum

#1 2011-08-16 07:19:35

dividing polynomials

#2 2011-08-16 08:12:33

Re: dividing polynomials

#3 2011-08-17 01:08:36

Re: dividing polynomials

#4 2011-08-17 01:54:14

Re: dividing polynomials

#5 2011-08-17 02:21:12

Re: dividing polynomials

#6 2011-08-17 03:02:43

Re: dividing polynomials

#7 2011-08-17 03:04:51

Re: dividing polynomials

#8 2011-08-17 03:07:26

Re: dividing polynomials

#9 2011-08-17 03:13:13

Re: dividing polynomials

#10 2011-08-17 03:14:30

Re: dividing polynomials

#11 2011-08-17 03:18:09

Re: dividing polynomials

#12 2011-08-17 03:23:58

Re: dividing polynomials

#13 2011-08-17 03:45:42

Re: dividing polynomials

#14 2011-08-17 04:12:58

Re: dividing polynomials

#15 2011-08-17 08:11:16

Re: dividing polynomials

#16 2011-08-17 08:15:44

Re: dividing polynomials

#17 2011-08-17 08:33:28

Re: dividing polynomials

#18 2011-08-17 10:05:20

Re: dividing polynomials

#19 2011-08-17 10:37:47

Re: dividing polynomials

#20 2011-08-17 11:13:51

Re: dividing polynomials

#21 2011-08-17 13:44:48

Re: dividing polynomials

#22 2011-08-17 22:34:31

Re: dividing polynomials

#23 2011-08-17 22:38:13

Re: dividing polynomials

#24 2011-08-17 22:40:37

Re: dividing polynomials

#25 2011-08-17 22:41:52

Re: dividing polynomials

Board footer