Question

What are the factors of X^2+16y^2

Answer 1

That can't be the only information you were given, there's really nothing that can be determined form that little information.

Answer 2

it says to identify the factors of x^2+16y^2

Answer 3

OH! Omg, i completely messed up when looking at that, okay, so have you learned what perfect squares are yet?

Answer 4

yes but I couldn't figure it out do you know what the answer is

Answer 5

please help

Answer 6

Okay, so when looking for perfect squares you look for something like this\[a^2x^2+b^2y^2\]you try to find a coefficient for each variable that is a perfect square, notice that for you, you can replace the a and b in that formula by\[a^2=1,b^2=16\]now that you have that information, you can simplify those to say that\[a=1,b=4\]now to factor this knowing those things, a perfect square looks like this\[(ax+by)(ax-by)\]now can you solve that question?

Answer 7

would the answer be (x+4y) (x-4y)?

Answer 8

That is completely correct! Great work, just remember, that if you are ever in doubt, do the FOIL method to check if you get back to the original equation, the FOIL method being, you multiply the FIRST parts, then you multiply the OUTER parts, then the INNER parts, then the LAST parts, adding and subtracting them as their coefficients tell you to

Answer 9

Great thank you I have another question what are the factors x^2-8x-12

Answer 10

Alrighty, so for non perfect square factoring, which means when you see a number with a variable of power 1, the way i do it is look at the the number without the variable, in this case\[12\]and ask myself, what factors of that can add or subtract together to get the coefficient of that middle part? well, 6x2=12, and 6+2=8, but now we have a problem, since that 8 is negative, we'd have to do -6+-2=-8 which works, but then -6x-2=12 which is not the -12 that we want, so we have to resort to the quadratic equation which looks like this,\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]when you have a quadratic function of the form\[ax^2+bx+c=0\]yes, there is an implied 0 on the other side when looking for factors, so can you plug your numbers into that now? i know it looks bad haha, but that's how you do it

Answer 11

So in factored form what would it be (x+6) (x-2)?

Answer 12

like i said, you can't just simply factor this problem, you have to plug it into the Quadratic Equation because it doesn't actually factor nicely into something like that

Answer 13

So it would be prime because all the other answers are in that form

Answer 14

Answer? type out those answers please, i'd like to see them before i continue

Answer 15

A) (x-6)(x+2)
B)(x+3)(x-4)
C) Prime
D) (x+6)(x-2)

Answer 16

Oh my, that terminology "prime" isn't something i ever learned, sorry for not understanding, but yes, after looking that up, the answer is C) that the result is "prime" because the factors don't add up 8

Answer 17

Thanks so much that's what I thought but I was not sure do you have time to help me with some more questions?

Answer 18

I can help you yeah, but i'd much prefer if you try them and get an answer, and post your result along with the question so i can see if i can figure out where you went wrong as well.

Answer 19

I would do that but I don't even know where to start haha

Answer 20

Ha, alright then, go ahead and ask, this time i'll try and give you a complete way to solve the rest

Answer 21

which shows 63^2-37^2 being evaluated using the difference of perfect squares method and let me know if you need the answer choices

Answer 22

Sooo, remember that for perfect squares, what you are looking for is an equation of the form\[a^2x^2-b^2y^2\]where any of those letters can be any number, or variable, they just have to be square able, well notice that\[(63)^2=(9x7)^2=9^2*7^2\]hey, that looks like a=9 and x=7 to me, now look at the other piece\[(37)^2=(37)(37)=(\sqrt{37})^2(\sqrt{37})^2\]that one required a bit of manipulating, it's still the same number in the end, and hey, that looks like b=sqrt(37) and y=sqrt(37), convert this into the perfect square equation to get\[(aX+bY)(ax-by)=(9*7+\sqrt{37})(9*7-\sqrt{37})=(63+\sqrt{37})(63-\sqrt{37})\]

Answer 23

i need to know... your very first question, was there really a + in between the two variables? because if so, then that isn't even a perfect square, and i'm kicking myself because i again, misread the question

Answer 24

yeah I got it wrong