HELP! SIMPLIFYING stuff. please please help.?

Question

:) kay thanks.
so i've got this test but i dont really get it..please help. (the stuff doesn't work when i type it sideways.)

*** each STAR (*) represents squared.

8a*+2a-15
_________
4a*-4a-15

it says i need to simplify that. i dont get it though.

oo and also simplifying the following rational expressions

x*-2x
_____ X
x*-x-2


x*-1
____
x*-x


....



2
___ +
x-4


5
____
x*-16

and...

2x
____ -
2x+1

4x
____ -
2x+3


5
________
4x*+8x+3

I can understand that it is hard to read vertically but...please try? thanks.

Answer 1

(8a^2+2a-15)/(4a^2-4a-15)

factor the top to (2a+3)(4a-5)
and the bottom to (2a+3)(2a-5)

so (2a+3)(4a-5)/((2a+3)(2a-5))

= (4a-5)/(2a-5)

Answer 2

Problem 1
(8a^2 + 2a - 15)/(4a^2 - 4a - 15)
factorize both numerator (NR) and denominator(DR)
NR = 8a^2 + 2a - 15
= 8a^2 + 12a - 10a - 15
= 4a(2a + 3) - 5(2a +3)
= (4a - 5)(2a + 3)
DR = 4a^2 - 4a - 15
= 4a^2 - 10a + 6a - 15
= 2a(2a - 5) + 3(2a - 15)
=(2a + 3)(2a - 5)
NR/DR = [(4a - 5)(2a + 3)]/[(2a + 3)(2a - 5)]
= (4a -5)/(2a - 5)

Problem 2
[(x^2 - 2x)/(x^2 - x - 2)] *[(x^2 - 1)/(x^2 - x)]
NR of first fraction = x^2 - 2x = x(x -2)
DR of first fraction = x^2 - x - 2
= x^2 - 2x + x -2
= x(x - 2) + 1(x - 2)
=(x - 2)(x + 1)
NR/DR of first fraction = x(x - 2)/(x - 2)(x + 1)
= x/(x + 1)
NR of second fraction = x^2 - 1
= (x + 1)(x - 1) (since a^2 - b^2 = (a + b)(a - b))
DR of second fraction = x^2 - x = x(x - 1)
NR/DR of second fraction = (x + 1)(x - 1)/(x)(x - 1)
=(x + 1)/x
So [(x^2 - 2x)/(x^2 - x - 2)] *[(x^2 - 1)/(x^2 - x)] =
[(x)/(x+1)] * [(x + 1)/x] = 1

Problem 3
[2/(x - 4)] + [ 5/(x^2 - 16)]
DR of second fraction = x^2 - 16
( using formula a^2 - b^2 =(a + b)(a- b)
x^2 - 16 = (x + 4)(x - 4)
[2/(x - 4)] + [ 5/(x^2 - 16)] =
[2/(x - 4)] + [ 5/(x + 4)(x - 4)] =
[2(x + 4) + 5]/[(x +4)(x -4)] =
(2x + 8 + 5)/(x + 4)(x - 4) =
(2x + 13)/(x + 4)(x -4)

Poblem 4
[2x/(2x + 1)] - [4x/(2x + 3)] - [5/(4x^2 + 8x + 3)]
DR of third fraction = 4x^2 + 8x + 3
= 4x^2 + 6x + 2x + 3
=2x(2x + 3) + 1(2x + 3)
=(2x + 1)(2x + 3)
So [2x/(2x + 1)] - [4x/(2x + 3)] - [5/(4x^2 + 8x + 3)] =
[2x/(2x + 1)] - [4x/(2x + 3)] - [5/(2x + 1)(2x + 3)]=
[(2x)(2x + 3) - (4x)(2x + 1) - 5]/(2x + 1)(2x + 3) =
[4x^2 + 6x - 8x^2 - 4x - 5]/(2x + 1)(2x + 3) =
[- 4x^2 + 2x - 5]/(2x + 1)(2x + 3)

Answer 3

Use the F. O. I. L. method

That's all you get from me, didn't do my kid's homework for them either.

Answer 4

i know how to do these but its really difficult typing it on the comp especially explaining