Question

How about 4x^2 + 81- 36x

Answer 1

what exactly is the question

Answer 2

i dont know how to do the problem, and i want someone to walk me throught the steps so i can understand the futture questions. 4x^2+81-36x

Answer 3

When factoring any quadratic, the goal is always the same. Find two numbers (x and y) that add to get b, but multiply to get ac.

x + y = b
xy = ac

Answer 4

In this case, we need to find:

x + y = -36
xy = 324

Answer 5

The only two numbers that work is

x = -18
y = -18

Answer 6

Anytime both x and y are the same, it means you will have a perfect square.

Answer 7

4x^2 - 36x + 81
4x^2 - 18x - 18x + 81
2x(2x - 9) - 9(2x - 9)
(2x - 9)(2x - 9)
(2x - 9)^2

Answer 8

alright. so how about 4y^2 +16y+16
How would that work..

Answer 9

It works the same way:

x + y = 16
xy = 64

Find x and y

Answer 10

I'm pretty sure you can find the two numbers

Answer 11

It doesn't require doing any actual Algebra. All you do is ask yourself, "What two numbers add to get 16, yet multiply to get 64?".

Answer 12

In this case, the two numbers are the same.

Answer 13

I'm not going to your homework for you. I'll show you how to do one problem. But, don't think I'm going to show you how to do every single problem. You need to figure this one out. The steps are exactly the same as the previous problem.

Answer 14

i dont expect you to do it for me lol. i actually need to learn it. im not in hs like a bunch of the people on there. i am in nursing school and personally do not want to have to repay for the class.

Answer 15

Okay, so did you find the two numbers yet?

Answer 16

if you are having trouble, a simple way of solving this is substitution.

x + y = 16
xy = 64

or in other words

y = 16 - x
x(16 - x) = 64

further simplified...

16x - x^2 = 64

x^2 - 16x + 64 = 0

then you can factor and solve

Answer 17

which is a much more simple polynomial to answer than the previous one

Answer 18

Funny thing is, she would still need to find the same two numbers either way.

Answer 19

i know i realized that after i posted it :D

but it's still a more simple polynomial to look at, at least in terms of co-efficients

Answer 20

ok. i think i got it.

Answer 21

4y^2 + 16y +16
4(y^2 + 4y + 4)
4(y^2+ 2y + 2y+ 2)
4 [y(y+2) + 2(y+2)]
=4(y+2)^2

Answer 22

right? yes no? lol ive always been terrible with numbers :/

Answer 23

Good job

Answer 24

sheesh. glad i dont ddeal with this in my profession ha

Answer 25

You have to think of math as "playing with numbers", rather than "a chore".

Answer 26

its just too difficult for me to understand. anyting else you can put in front of me and i get it right away, and sure certain parts of math i can too, but when it comes to this i just...ugggg. wanna curl up in a ball and cry

Answer 27

Most of the time, students that don't get it are usually those who are not very good with seeing the relationships between addition and multiplication. Either that or they don't have their multiplication tables well memorized.

Answer 28

This concept is definitely more "mental math" compared to other algebra concepts.

Answer 29

my tables are fine...and maybe i do have a problem with my adding and multiplying and i just dont realize it?
cuz ive never gotten the factoring polynomials things. not now, not in hs

Answer 30

I'm sorry to hear that.

Answer 31

hey. only gotta deal with this until march so i think i can stick it out

Answer 32

If you have difficulty with "finding the two numbers that add to get one number, but multiply to get another", then your problem is not with factoring, but with the multiplication/addition relationships.

Answer 33

yeah. thats what i thought.

Answer 34

well thank you! i appreciate it very much!! back to other homework. Happy Holidays!