Question

Factor completely p^4 – 18p^2 + 81

Answer 1

let q = p^2 (so q^2 is (p^2)^2= p^4 )

you have
q^2 -18q +81

can you factor this quadratic ?

Answer 2

yes

Answer 3

what do you get ?

Answer 4

(q-3)(q+3)

Answer 5

?

Answer 6

look at the 81:
list pairs that multiply to give 81:
1,81
3,27
9,9

which pair will give 18 ? (the middle number , from -18q)

Answer 7

my bad i meant to put 9

Answer 8

ok, now what about the signs? the + sign on +81 means both roots are the same sign

the - on the -18x means the largest root is negative

Answer 9

so it would be (p-9)(p+9)

Answer 10

no.... first it is still q (not p)
second , the signs *are the same*

Answer 11

okay, what would i do now?

Answer 12

what do you have so far ?

Answer 13

(q-9)(q-9)?

Answer 14

yes... you can check by multiplying it out to see if you get the original... which you will.

now we set q= p^2 . replace q with p^2. you get
(p^2 - 9) ( p^2-9)

can you factor (p^2-9) ?

Answer 15

using "difference of squares"
9 is 3*3 so this is the same as (p^2 - 3^2)

Answer 16

(p-3)(p+3), btw thanks man for teaching me not giving an answer

Answer 17

yes, and we have two (p^2 - 3^2) :
(p^2 - 3^2)(p^2 - 3^2)
and you get
(p-3)(p+3)(p-3)(p+3)

you can write it a little more simply.... (p-3)(p-3)= (p-3)^2
and the same for (p+3)(p+3)

Answer 18

either way is the answer, which one to use depends if you have multiple choice
(p-3)(p+3)(p-3)(p+3)
or
(p-3)^2 (p+3)^2

Answer 19

thanks