Question

can someone help me with this question 10y^2+9y-9 using grouping ac= and b= i have know clue what to do on this can someone help me with this please

Answer 1

Are you looking to factor this?

Answer 2

If so the factored form is: (2 y+3) (5 y-3)

Answer 3

no just the ac= and that is what im stuck on to find the ac it that the number + the number or the number * the number

Answer 4

ax^2 + bx + c

Answer 5

ok so will a = 10y2 and c = -9

Answer 6

The coefficients equal numbers only. Don't ever include the variables

Answer 7

You must be confused again

Answer 8

yes i am

Answer 9

Find two numbers, m and n, that multiply to get ac, yet add to get b:

mn = ac
m + n = b

Answer 10

ok so do i have to solve the problem like we did yesterday or do i just pull the 2 numbers out

Answer 11

You always have to solve it like that

Answer 12

ok

Answer 13

so will it be 9 and 1

Answer 14

mn = -90
m + n = 9

Answer 15

Funny thing is, we did something similar to this before using the exact same numbers.

Answer 16

I see that you have forgotten them

Answer 17

You only memorized your multiplication tables up to twelves, I bet.

Answer 18

im sorry im bad at math not really i was cheating using a calculator

Answer 19

I'm sorry to hear that.

Answer 20

its all good

Answer 21

15 - 6 = ?

Answer 22

9

Answer 23

Yep, so now you have what you need. Post your factorization here.

Answer 24

10y^2+15y-9

Answer 25

I thought you were getting the hang of it, but I see that you are not.

Answer 26

ok so its 10y^2+15y-6

Answer 27

You basically changed it to a different trinomial.

Answer 28

o dang it well i thought i was getting it yesterday but i guess not

Answer 29

and i have a test Monday on all of this stuff

Answer 30

If you didn't write down the steps I showed you yesterday...

Answer 31

i did wright them down

Answer 32

10y^2+9y-9
10y^2 + 15y - 6y - 9
5y(2y + 3) -3(2y - 3)
(2y + 3)(5y - 3)

Answer 33

Notice that you only need to change the middle term, then you can start factoring:

9y = 15y - 6y

Answer 34

o ok i forgot to put the 15 and the 6 together dang it

Answer 35

i keep for getting that part

Answer 36

Here's a page where you can practice. Scroll down.

http://www.themathpage.com/alg/factoring-trinomials.htm

Answer 37

ok thank you

Answer 38

do you think you can help me more please

Answer 39

12x^2-13x-4

Answer 40

You're supposed to be practicing on your own.

Answer 41

Can you at least come up with the equation for m and n?

Answer 42

i can try to hold on

Answer 43

ok so do have to change 12 because i cant find a number to make -13 and -4

Answer 44

You're supposed to be solving this:

mn = ac
m + n = b

In other words

mn = (12)(-4)
m + n = -13

Answer 45

I usually multiply the a and the c and express it like this:

mn = -48
m + n = -13

Answer 46

You won't be able to get stuck like this on the test, so you better learn quick. Anyway, we have plenty to practice. Try doing the next step .

Answer 47

This process of figuring out the m and the n is a mental process. You're supposed to ask yourself, "What are two numbers that multiply to get -48, yet add to get -13?".

Answer 48

I just write it out explicitly so that you can see more easily what needs to be done.

Answer 49

ok so its -16 and 3

Answer 50

then the next step is 12x^2-16x-3x-4

Answer 51

Did you use your calc to get that?

Answer 52

i just divided 48 and 3and got it that way

Answer 53

I see. Good job. Well...finish it. Wait. You have to make sure the middle two terms add to get -13. Double check your latest expression.

Answer 54

12x^2-16x+3x-4

Answer 55

There we go. If you figure out this next part, then you'll be well on your way.

Answer 56

You have to factor the first two terms; then factor the last two terms. Then factor what's common to both.

Answer 57

12x(x-16)+3(x-16)
(x+16)(-3)

Answer 58

Sigh

Answer 59

12x^2-16x+3x-4
4x(3x - 4) + 1(3x - 4)
(3x - 4)(4x + 1)

Answer 60

Study that.

Answer 61

Just a quick question....Do you know how to multiply binomials?

Answer 62

not really

Answer 63

That's where the problem lies.

Answer 64

The intuition necessary to be able to factor trinomials comes from mastery of mulitplying binomials. You can't learn how to factor trinomials without first learning how to multiply binomials.

Answer 65

my last math teacher didnt teach a lot he talked about other stuff like his family and the moon and stuff like that and now im having problems with this math class

Answer 66

Well, let's learn how to multiply binomials real quick.

Answer 67

Do you remember the distributive rule?

Answer 68

Or rather, the distributive property?

Answer 69

@jen89, are you still there?

Answer 70

yes im still here sorry i had to do something really fast

Answer 71

i dont realy remember that at all

Answer 72

(a + c)b = ab + bc

Answer 73

That's distributive rule.

Answer 74

When multiplying binomials, you do a similar technique:

(x + 3)(x + 4) = x(x + 4) + 3(x + 4)

Answer 75

That's the first step

Answer 76

x(x + 4) + 3(x + 4) = x^2 + 4x + 3x + 12

Answer 77

x^2 + 4x + 3x + 12 = x^2 + 7x + 12

Answer 78

after multiplying two binomials, you end up with a trinomial

Thus (x + 3)(x + 4) = x^2 + 7x + 12

Answer 79

ok i wrote the steps down

Answer 80

Well, I was showing you step by step the process. Now, here's how I would usually introduce the steps to a student...in a more straight forward, but simpler form:

(x + 3)(x + 4)
= x(x + 4) + 3(x + 4)
= x^2 + 4x + 3x + 12
= x^2 + 7x + 12

Answer 81

Try to understand those steps.

Answer 82

If you look closely, you'll notice that it's actually the reverse of factoring.

Answer 83

Multiplying:

(x + 3)(x + 5)
= x(x + 5) + 3(x + 5)
= x^2 + 5x + 3x + 15
= x^2 + 8x + 15

Factoring:

x^2 + 8x + 15
= x^2 + 5x + 3x + 15
= x(x + 5) + 3(x + 5)
= (x + 5)(x + 3)

Answer 84

Multiplying binomials and factoring trinomials have an inverse relationship.

Answer 85

ok i see that now when you put both together

Answer 86

That is what you must master.

Answer 87

ok

Answer 88

is there any way i can learn all of this before monday

Answer 89

Yes, you can. all you need to do is study it. It is a template to use for other factoring methods. Basically, you just familiarize yourself with it. Try to think of it like getting to know someone you just met for the first time. When you meet someone for the first time and you're really interested in them, you try to learn as much as possible about them. You want to be a good friend to them and you may even want to have a mutual relationship with them. You should want to do something similar with learning this.

Answer 90

ok thank you i will do my best learning all of this