At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
what exactly is the question
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
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
In this case, we need to find: x + y = -36 xy = 324
The only two numbers that work is x = -18 y = -18
Anytime both x and y are the same, it means you will have a perfect square.
4x^2 - 36x + 81 4x^2 - 18x - 18x + 81 2x(2x - 9) - 9(2x - 9) (2x - 9)(2x - 9) (2x - 9)^2
alright. so how about 4y^2 +16y+16 How would that work..
It works the same way: x + y = 16 xy = 64 Find x and y
I'm pretty sure you can find the two numbers
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?".
In this case, the two numbers are the same.
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.
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.
Okay, so did you find the two numbers yet?
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
which is a much more simple polynomial to answer than the previous one
Funny thing is, she would still need to find the same two numbers either way.
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
ok. i think i got it.
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
right? yes no? lol ive always been terrible with numbers :/
sheesh. glad i dont ddeal with this in my profession ha
You have to think of math as "playing with numbers", rather than "a chore".
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
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.
This concept is definitely more "mental math" compared to other algebra concepts.
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
I'm sorry to hear that.
hey. only gotta deal with this until march so i think i can stick it out
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.
yeah. thats what i thought.
well thank you! i appreciate it very much!! back to other homework. Happy Holidays!