Question

What is the least common denominator for the equation (4p)/(3p^2-27) = (p+1)/(2p^2+10p+12)

Answer 1

\[\frac{ 4p }{ 3p ^{2-27} }= \frac{ p +1 }{ 2p ^{2}+10p +12 }\]

Is this yoru equation?

Answer 2

the 27 isn't an exponent.

Answer 3

or is the minus 27 part of the binomial and not part of the exponent?

Answer 4

All you do is factor both out

Answer 5

part of the binomial

Answer 6

\[\frac{ 4p }{ 3p ^{2}-27 }=\frac{ p +1 }{ 2p ^{2}+10p +12 }\]

Answer 7

So first, you factor the denominators of each rational expression.

Answer 8

Okay.
so 3(p^2-9) and ...Idk how to factor trinomials

Answer 9

\[\frac{ 4p }{ 3\left( p -3 \right)\left( p +3 \right) }=\frac{ p +1 }{2\left( p +2 \right)\left( p +3 \right) }\]

Answer 10

ok, before that on the left, just factor the 3, you got\[3\left( p ^{2}-9 \right) \]

that is called the difference of squares, anythign where you can do a square root of the first term and the second, with a minus sign, factors to (x+y)(x-y) where x is the square root of the first term, and y is the square root of the 2nd term, so x is the square root of x^2, and 3 is the square root of 9

Answer 11

Ohhh.

Answer 12

The way to factor a trinomial of the form x^2 +bx +c, you have to find factors of C that add to get bx, so basically

\[2p ^{2}+10p +12\]
\[2\left( p ^{2}+5p +6 \right)\] Factor out the 2

Now, you need to find 2 factors of 6, so they multiply to get 6, and add up to get 5
Those are "2" and "3"
You take those two factors and add them to teh square root of the first term
So,
\[2\left( p +2 \right)\left( p +3 \right)\]

Answer 13

With me so far?

Answer 14

Yeah. Kinda

Answer 15

There's no real other way to explain how to factor a trinomial. That's just what you have to remember. You need to find 2 numbers that multiply to get the last term, and add up to get the middle term. It can help to make a table for bigger numbers, but for "6",
There's only 4 numbers that are divisible

1, 1X6=6
2, 2X3=6
3, 3X2=6
6, 6X1=6

So 1, 2, 3, and 6. Which two also add up to get 5? 3 and 2

Answer 16

So, now you have

\[\frac{ 4p }{ 3\left( p-3 \right)\left( p+3 \right) }=\frac{ p+1 }{ 2\left( p+2 \right)\left( p+3 \right) }\]

Answer 17

The (p+3) are mentioned twice, so only use one of them
You have (p+2)
You have (P-3)
You have 3
You have 2

OR

\[\left( 3 \right)\left( 2 \right)\left( p-3 \right)\left( p+3 \right)\left( P+2 \right)\]

Answer 18

You can just make the 3 and 2 into a 6, so the LCD is

\[6\left( p-3 \right)\left( p+3 \right)\left( p+2 \right)\]

Answer 19

Do you need to solve the equation too?

Answer 20

You'll want to practice factoring a whole bunch.

Factoring binomials
Factoring trinomials
Factoring quadranomials by grouping, etc.

You want to master factoring before you move onto anything else, because it basically needs to be 2nd nature to you before you move on to stuff like this.

Answer 21

Okay, thank you