Question

simplify each rational expression state any restrictions on the variable.

Answer 1

\[\frac{ 4x^2-2x }{ x^2+5x+4} \div \frac{ 2x }{ x^2+2x+1 }\]

Answer 2

help @whpalmer4

Answer 3

Factor everything in sight before you do the division, and you'll get some significant simplification. Then proceed as always...

Answer 4

ok im not great at factoring can you help me with that part

Answer 5

Okay, can you spot any common factors in \[4x^2-2x\]?

Answer 6

um 2 can go into 4 and x can go into x^2

Answer 7

so, factored, that would be?

Answer 8

um i dont know ....... would it be (2+x)(2x)

Answer 9

nope that was not it

Answer 10

ok i think i got it (2)(2x-x)

Answer 11

\[4x^2-2x = 2(2x^2-x) = 2x(2x-1)\]

Answer 12

ok how do u \[x^2+5x+4\]

Answer 13

Okay, remember when multiply two binomials\[(x+a)(x+b) = x*x+a*x+b*x + a*b\]\[=x^2+(a+b)x + ab\]
\[x^2+5x+4\]Those match up if \[a+b=5\]\[a*b=5\]Can you think of two numbers for which that is true?

Answer 14

sorry, that last line should be \[a*b=4\]!!!

Answer 15

a=2 b=2 which is 4

Answer 16

does 2+2 =5 in your world? :-)

Answer 17

yes it should

Answer 18

well, maybe it should, but it doesn't.

how about 1 and 4?

1*4 = 4
1+4 = 5

\[(x+1)(x+4) = x*x + 1*x + 4*x + 1*4 = x^2 + 5x + 4\]

Answer 19

ok so the new equation will look like
\[\frac{ 2x(2x-1) }{ (x+1)(x+4) } \div \frac{ 2x }{ (x+1)(x+1) }\]

Answer 20

right. and that becomes a multiplication with the \(2x/(x+1)^2\) inverted:

\[\frac{2x(2x-1)}{(x+1)(x+4)}*\frac{(x+1)(x+1)}{2x}\]

Answer 21

simplified it is
\[\frac{ 2x-1 }{ (x+4)(x+1) }\]

Answer 22

i fixed it

Answer 23

uh, no, not quite...

\[\frac{\cancel{2x}(2x-1)}{\cancel{(x+1)}(x+4)}*\frac{(x+1)\cancel{(x+1)}}{\cancel{2x}}=\frac{(2x-1)(x+1)}{(x+4)} \]

Answer 24

to find the restrictions on the variable. dont we need to find the zeros?

Answer 25

Yes, what will make the denominator = 0?

Answer 26

x=-4

Answer 27

right??

Answer 28

are the other ones x=-4, x=-1, x=1

Answer 29

not 1 but 1/2

Answer 30

yes, x=-4 is the restriction on the variable — \(x\ne-4\)

Answer 31

the others are zeros, x=-4 is a pole

Answer 32

what??? i thought the restriction are the zeros

Answer 33

No, zeros are where the thing equals 0. There's nothing wrong with that!

Restrictions are where you are dividing by 0, which isn't allowed.

Answer 34

ok i think i get it but when i use -1 it comes up on the bottom as 0. so it that also a restriction

Answer 35

ok so the restrictions are x=-1 & x=-4

Answer 36

ah, yes, because the original has (x+1) on the bottom, so that's a restriction even though it doesn't appear in the simplified fraction

Answer 37

sorry to leave you mid-problem, but I have to go!

Answer 38

ok so i have answer thank you for you time do you think you can help again if i post a new question?

Answer 39

go ahead and post, tag me, I'll have a look when I'm able. Hopefully someone else can help you before I'm back.

Answer 40

ok thanks again

Answer 41

you're welcome.