Question

find the least common denominator for these two rational expressions. Please help, will give medals for correct answers!!!

Answer 1

here's the problem...I know that you have to FOIL it but after that step, i'm stuck.

Answer 2

Can you factor the polynomials?

Answer 3

What wio said will be a great start

Answer 4

yep i can do that step...i just need help after wards...

Answer 5

The LCD is given by the product divided by the GCD

Answer 6

What do you get when you factor?

Answer 7

i'm having trouble factoring this...this is what i have so far... (y+1)(y+1)

Answer 8

Okay, so why don't we start with \(y^2+4y+4\)?

Answer 9

The factors of \(4\), are going to be \(1,2,4\) as well as their negative counter parts, so they make for good guesses.

If we wanted to avoid guessing all together, we could use the quadratic formula. Do you know it?

Answer 10

I don't know it...i'd rather avoid just guessing:)

Answer 11

So how were you taught to factor things like this?

Answer 12

i googled it, i knew FOIL but that's all, i was mostly guessing.

Answer 13

Okay, well if you foil:\[
(x+a)(x+b) = x^2+(a+b)x+ab
\]This means we have: \[
a+b=4\\
ab=4
\]Since those were our coefficients.

Answer 14

If I may add, when you are factoring a quadratic with no coefficient(number in front of) the \(x^2\) term, you want to find two numbers that add to b and multiply to c. \[x^2+bx+c\]
So you look at the factors of c, then a quick addition to see which yield b.

Answer 15

and you have your two numbers to plug in. BUT as wio said, you can always do a quadratic :)

Answer 16

ok coefficient of 1**

Answer 17

I'm stuck after getting the FOIL parts....

Answer 18

So one guess would be \(1\times 4=4\), but since \(1+4=5\neq 4\), those won't work.
Next guess would be \(2\times2=4\), and \(2+2=4\), therefore\[
(y+2)(y+2)=y^2+4y+4
\]

Answer 19

It's really like unFOILing

Answer 20

@specek18 Are you able to understand what I wrote so far?

Answer 21

yes:) so you were basically just eliminating what doesnt work?

Answer 22

Yeah

Answer 23

what is the next step?

Answer 24

Now, for the second one, we need to use factors of \(8\) to try out.

Answer 25

y^2+4y+4....how do you use the factors of 8?

Answer 26

No, we already did that one. We know:\[
(y+2)(y+2)=y^2+4y+4
\]

Answer 27

Now we are doing \(y^2+6y+8\).

Answer 28

oh okay, we're on the second. I'm sorry! that's why i was confused.

Answer 29

We want \(a+b=6\) and \(ab=8\). That is why we are factoring \(8\)... to find \(ab\) values.

Answer 30

2, and 4?

Answer 31

Yep, those work. Therefore: \[
(y+2)(y+4) = y^2+6y+8
\]

Answer 32

Next step is to find the GCD of the two.

Answer 33

What factors do they both have in common?

Answer 34

they both have the y^2 and the 4 is a factor of 8?

Answer 35

They both have \(y+2\) as a factor...

Answer 36

okay...that makes sense. sorry i have sooo much trouble with math.

Answer 37

So the GCF of \((y+2)(y+2)\) and \((y+2)(y+4)\) is \(y+2\).
The LCD will just be the product divided by the GCF.\[
\frac{\cancel{(y+2)}(y+2)(y+2)(y+4)}{\cancel{y+2}} = (y+2)(y+2)(y+4)
\]

Answer 38

I should enter the (y+2) twice? i see where you cancelled things out,,...just don't know about the y+2...

Answer 39

So the GCF of \(\color{blue}{(y+2)(y+2)}\) and \(\color{green}{(y+2)(y+4)}\) is \(\color{red}{y+2}\).
The LCD will just be the product divided by the GCF.\[
\frac{\cancel{\color{blue}{(y+2)}}\color{blue}{(y+2)}\color{green}{(y+2)(y+4)}}{\cancel{\color{red}{y+2}}} = (y+2)(y+2)(y+4)
\]

Answer 40

The top part is what happens when you multiply both of them, the bottom part is the GCF

Answer 41

it was correct!!!!!!!

Answer 42

ready for another? lol

Answer 43

You'll have to ask another question.