anonymous
  • anonymous
How about 4x^2 + 81- 36x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what exactly is the question
anonymous
  • anonymous
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
Hero
  • Hero
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

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Hero
  • Hero
In this case, we need to find: x + y = -36 xy = 324
Hero
  • Hero
The only two numbers that work is x = -18 y = -18
Hero
  • Hero
Anytime both x and y are the same, it means you will have a perfect square.
Hero
  • Hero
4x^2 - 36x + 81 4x^2 - 18x - 18x + 81 2x(2x - 9) - 9(2x - 9) (2x - 9)(2x - 9) (2x - 9)^2
anonymous
  • anonymous
alright. so how about 4y^2 +16y+16 How would that work..
Hero
  • Hero
It works the same way: x + y = 16 xy = 64 Find x and y
Hero
  • Hero
I'm pretty sure you can find the two numbers
Hero
  • Hero
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?".
Hero
  • Hero
In this case, the two numbers are the same.
Hero
  • Hero
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.
anonymous
  • anonymous
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.
Hero
  • Hero
Okay, so did you find the two numbers yet?
anonymous
  • anonymous
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
anonymous
  • anonymous
which is a much more simple polynomial to answer than the previous one
Hero
  • Hero
Funny thing is, she would still need to find the same two numbers either way.
anonymous
  • anonymous
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
anonymous
  • anonymous
ok. i think i got it.
anonymous
  • anonymous
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
anonymous
  • anonymous
right? yes no? lol ive always been terrible with numbers :/
Hero
  • Hero
Good job
anonymous
  • anonymous
sheesh. glad i dont ddeal with this in my profession ha
Hero
  • Hero
You have to think of math as "playing with numbers", rather than "a chore".
anonymous
  • anonymous
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
Hero
  • Hero
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.
Hero
  • Hero
This concept is definitely more "mental math" compared to other algebra concepts.
anonymous
  • anonymous
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
Hero
  • Hero
I'm sorry to hear that.
anonymous
  • anonymous
hey. only gotta deal with this until march so i think i can stick it out
Hero
  • Hero
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.
anonymous
  • anonymous
yeah. thats what i thought.
anonymous
  • anonymous
well thank you! i appreciate it very much!! back to other homework. Happy Holidays!

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