Question

Factorising:
i don't understand the method to answer the following...
simplify (3p^2-21p)/(3p^2-20p-7)
(i will put the actual equations below)

Answer 1

\[\frac{ 3p ^{2}-21p }{ 3p ^{2} -20p-7}\]

i know that the top of the fraction would be:
\[3p(p-7)\]
but what is the bottom?

Answer 2

The denominator factors as (3p+1)(p-7)

Answer 3

okay thanks, how do you find this out? that's just the part i'm struggling with :L

Answer 4

eg. how do you know it's (3p+1)(p-7) rather than (3p-7)(p+1)?

Answer 5

(3p-7)(p+1) does not multiply back to the correct middle term of -20p,

-7 *p +1*(3p) = -7p + 3p = -4p and not -20p.

Answer 6

(3p-7)(p+1) does not equal (3p^2-20p-7)

Answer 7

(3p+1)(p-7) = (3p^2-20p-7)

Answer 8

ah okayy thanks,
so another question:
\[3x ^{2}+8x-3\]

Answer 9

Here's a technique for factoring trinomials I wrote up based on how I learned to factor by busting the "b." You might want to look over it later.

Busting the "b"

4 x² -1x + 18

a = 4 and c = 18 and b = -1.

The task on "bust the b" is to find numbers that mutiply to a*c AND add to b.

In this case, numbers that multiply to 4*18 and add to -1.

14 * 18 breaks aparts into the product of prime factors : 2*2*2*3*3

Look at those five factor of 14 * 18 and try to put them in the form of 2 numbers that differ by -1.

2*2*2 and 3*3 appear to differ by 1. That would be 8 and 9.

"b" has been "busted into 8 and - 9 whose sum is -1 which equals "b" in this problem.

4x - x+ 18 = 4 x² + 8 x - 9 x - 1.

The "b" has been busted.

Factoring by grouping comes next.

4 x² + 8 x - 9 x - 18 =

4x ( x + 2) - 9 ( x + 2 ) = --> On this step, (x + 2) is the common factor of the expression.

[ (x + 2) ] ( 4x - 9 ) =

( x + 2 ) ( 4x - 9)

Answer 10

Did you finish this problem:

simplify (3p^2-21p)/(3p^2-20p-7)

Answer 11

If so, what did you get for the answer?

Answer 12

Is \[3x ^{2}+8x-3\] = 3x^2 + 8x - 3. The math processor is not working on the site or on my computer.

Answer 13

\[ \frac{ 3p }{ 3p+1 }\]

Answer 14

Click on the attachment to see what I see when the math processor is not working. :)

Answer 15

ah i get you... well basically it's: 3p/(3p+1)

Answer 16

3p/(3p+1) is the answer I got.

Answer 17

What are we doing here, factoring?

\[3x ^{2}+8x-3\]

Answer 18

Look for two numbers that multiply to -9 and add to -8.

Answer 19

@Directrix i was just about to explain that, how to factor trinomials when a>1 :P

I'll just link to my explanation, but yours is already here... but i do it slightly differently.

http://www.wyzant.com/answers/4707/explain_how_to_factor_the_following_trinomials_forms_x_bx_c

Answer 20

Thanks for your help :) i will read it all through, if theres any more questions i'll pop them on here, tah :)

Answer 21

While we are here, we can factor 3x^2 + 8x -3 if you have time.

Answer 22

ah yeh thanks, so to start off: would it be (3x ?)(x ?)

Answer 23

First find factors of -9 (since 3*(-3) = -9) that add up to +8.

Answer 24

3x^2 + 8x -3 = 3x^2 + 9x - 1x -3 = 3x(x + 3) -1 (x+3)= (x + 3)(3x-1).
It is factoring by grouping although bust the b comes from writing the 8x as the sum of 9x and -x.

Answer 25

the busting the b method seems way to hard, im doing gcse's year 11

Answer 26

It's just a slightly more organized technique than guess and check but that doesn't mean that busting the b is better than any other method. Whatever works for the learner is best, I think.

Answer 27

okay, well i will try a couple more questions.
how about if its not a trinomial, eg 9x^2-1

Answer 28

The way i do it is similar to "busting the b", but w/o the grouping.

factors of -9 that add up to 8 are -1 and 9, so i write:

(3x -1)(3x+9) .... but when we multiplied 3*(-3) earlier, we multiplied our expression by 3. So divide it by 3:

(3x -1)(3x+9)/3 = (3x-1)(x+3)

Done!

Answer 29

There are a few special factoring patterns that can be automated to help with the difference of two squares and perfect square trinomials.

Answer 30

oops noticed an error... fixed.

@katielong it'll get much easier with practice. I've seen students who really struggle with algebra use this method well.

eg. 4x^2 + 10x + 4 .... first multiply the first 4 by the last 4. 4x4 = 16. Factors of 16 that add up to the 10... 8 and 2. So write

(4x+8)(4x+2) ... and now don't forget to divide it all by the 4 we originally multiplied by:

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

Answer 31

question attempt:
3c^2-c-10

3*10 and +'s to make 1?

30=3*5*2?

so 3 and 2 have a difference of 1?

3c^2-c-10= 3c^2+3c-2c+1?

nope im stuck..... :S

Answer 32

@agent0smith i see now... i did it wrong too... let me try again...

Answer 33

Don't worry Katie, you'll get it with practice.

Well the way i do it is to just find factors... 3x-10=-30.

Factors of -30 that also add up to -1... +5 and -6 since 5-6=-1

(3c-6)(3c+5) now divide by the 3 we first multiplied by:

(3c-6)(3c+5)/3 = (c-2)(3x+5)

Answer 34

9x^2-1 = ?

Answer 35

what about say 2n^2-8?

Answer 36

^there you can first pull a factor of 2, then use diff. of two squares. \[2n^2-8 = 2(n^2 - 4) \]

Answer 37

ah i see thanks

Answer 38

so right,

simplify (n^2+6n-16)/(9x^2-1)

the top is (n+8)(n-2)

the bottom is 2(n-4)

so how do you do (n+8)(n-2)/2(n-4)?

Answer 39

wait... the bottom is 9x^2-1 which isn't 2(n-4). \[ 9x^2-1 = (3x)^2-1^2 \]now try using diff. of two squares on that. See directrix's pic above. I need sleep, will help later if you need it.

Answer 40

yeah thats fine thanks, haha its mid-day here :L

Answer 41

3:30 am here :P

btw that one doesn't appear to simplify easily....

(n+8)(n-2)/(9n^2-1)

diff of two squares on the bottom will give:

\[\frac{(n+8)(n-2) }{ (3n-1)(3n+1) }\] so there doesn't seem to be common factors...

Answer 42

hang on 2 questions are muddled here...
theres
1) \[\frac{ n^2+6n-16 }{ 2n^2-8 }\]
2) \[\frac{ 3x^2+8x-3 }{ 9x^2-1 }\]

Answer 43

ohhh... :)

yeah then the first one becomes (diff of two squares on denominator):

\[\frac{(n+8)(n-2)}{2(n^2-4) } = \frac{(n+8)(n-2)}{2(n+2)(n-2) } \] now try simplifying that :)

second (which we factored the top part of earlier) \[\frac{ 3x^2+8x-3 }{ 9x^2-1 } = \frac{ (3x-1 )(x+3) }{(3x-1)(3x+1) }\]

Answer 44

thanks i've got it now! go get some sleep haha, sorry fro keeping you up :L

Answer 45

You're welcome! And no prob :) gnight!