Question

Factor completely p4 – 16
last question i have i need help

Answer 1

help please

Answer 2

you posted this a second time ?

Answer 3

http://openstudy.com/#/updates/4efe3ffce4b01ad20b5302a6

Answer 4

@TuringTest has already given you a hint at how to solve this

Answer 5

yeah cuz i have no clue how to do this

Answer 6

do you understand this factorization:
\[a^2-b^2=(a+b)(a-b)\]

Answer 7

is a and b x and y?

Answer 8

like what would a and b be

Answer 9

(x^2-4)(x^2+4)

Answer 10

in your case:\[
a^2 = p^4\]\[b^2=16\]

Answer 11

(p^2-4)(p^2+4)

Answer 12

okay

Answer 13

done it

Answer 14

it can be factored again...
listen to what asnaseer is saying about the form in general

Answer 15

this is what is confusing me;
A.(p2 – 4)2
B.(p2 + 2)(p2 + 8)
C. (p2 + 4)(p + 4)(p – 4)
D. (p2 + 4)(p + 2)(p – 2)
at wasiqss your wrong it is something different cuz what u did i would of done to but its wrong

Answer 16

look at what you now have left:\[p^4-16=(p^2-4)(p^2+4)\]
notice the "form" of (p^2-4)

Answer 17

okay i have that written down

Answer 18

see if you can apply the same rule again to this

Answer 19

so i take the p^2-4 and apply it to the difference of squares?

Answer 20

so you know:\[a^2-b^2=(a+b)(a-b)\]
now:\[a^2=p^2\]\[b^2=4\]

Answer 21

yes - I think you are getting the hang of it now - well done

Answer 22

i got p^2-4^2=(p+4)(p-4)

Answer 23

bingo! - you're a genius!

Answer 24

oops - sorry candycane
you should have written:\[p^2-4=(p-2)(p+2)\]

Answer 25

remember \(b^2=4\), so \(b=2\)

Answer 26

so now if you put all the components together, what do you get?

Answer 27

(p2 + 4)(p + 2)(p – 2)

Answer 28

yay!
this time definitely right :)

Answer 29

thats it - you have the right answer - well done again.
the trick is in spotting when to use the difference of two squares factorization.

Answer 30

thank you soooo much! i definatly get it now

Answer 31

yw