algebra help please please?

Question

I have to divide this rational expression x^2 - 9x + 8 over 2x-4 divide x^2 -1 over 2x + 2 using [ ] to enclose the radicals can anyone help out thanks

Answer 1

Hi,

x^2.-.9x.+.8.....x^2 -1
---------------..÷..---------- Change to multiplying by the reciprocal.
..2x.-.4.............2x.+.2

x^2.-.9x.+.8.......2x.+.2
---------------..Х..---------- Now factor:
..2x.-.4...............x^2 -1

(x.-.8)(x.-.1).......2(x.+.1)
---------------..Х..---------------- Cancel 2, (x + 1), and (x-1)factors.
..2(x.-.2)............(x -1)(x + 1)

x.-.8
------- <=== That's the answer. .
x.-.2


I hope that helps!! :-)

Answer 2

Recall: When you divide fractions you take the reciprocal of the second fraction and then multiply and reduce. For example:

2/3 divided by 6/7
= 2/3 * 7/6
= 1/3 *7/3 (when you divide the numerator of the first fraction and the denominator of the second fraction by 2)
= 7/9

So, do the same with your problem:

(x^2 - 9x +8)/(2x - 4) divided by (x^2 - 1)/(2x + 2)
= (x^2 - 9x +8)/(2x - 4) * (2x + 2)/(x^2 - 1)

To reduce factor the numerators and denominators so you can see if any of the factors will divide out.

= [(x - 8)(x - 1)2(x+1)]/[2(x - 2)(x = 1)(x - 1)]

(x - 1) will divide out since it is in the numerator and the denominator.
(x + 1) will divide out since it is in the numerator and the denominator.
This leaves you with:

= (x - 8)/[2(x - 2)]

If you want you can distribute the 2 in the denominator so you'll get:

= (x - 8)/(2x - 4)

Note: There isn't any radicals in this problem so the [ ] are used as normal brackets.

Answer 3

x² - 9x + 8 = (x - 8).(x - 1) = A
2x - 4 = 2.(x - 2) = B

x² - 1 = (x - 1).(x + 1) = C
2x + 2 = 2.(x + 1) = D

Is question then [ A/B ] / [ C/ D] ?
If so this becomes AD / BC
= (x - 8).(x - 1).2.(x + 1) / 2.(x - 2).(x - 1).(x + 1)
= (x - 8) / (x -2)

Answer 4

is this two seperate problems? if not...

[(x^2-9x+8)/(2x-4)] / [(x^2-1)/(2x+2)] factor everything
=(x-1)(x-8)/2(x-2) * 2(x+1)/[(x-1)(x+1)] combine
=2(x-1)(x-8)(x+1)/2(x-2)(x-1)(x+1)] now, cancel out
=(x-8)/(x-2)

Answer 5

(x^2-9x+8/2x-4)/(x^2-1/2x+2)
=[(x-1)(x-8)/2(x-2)] x [2(x+1)/(x-1)(x+1)]
=(x-8)/(x-2)#

Answer 6

If these are two separate problems, then you need to learn long division, or synthetic division:
http://mathforum.org/library/drmath/view/56385.html