Question

@whpalmer4
Factor the polynomial.
-81 + a^4
Thank you! I will reward by giving you a medal! ^_^

Answer 1

Hey, wasn't I going to get an Oreo for the last one? :-)

Answer 2

AH this one seems too confusing...

Answer 3

Okay, \[-81+a^4\]Any ideas? How about we rewrite it in the other order, just to see if that makes anything jump out at us...

Answer 4

hmm? oh yes oreo...
okay, what's your address? i'll send it to you! (:
jk i don't wanna know..

Answer 5

Eat it for me, and enjoy it :-)

Answer 6

the greatest common factor that i can only think of is 1 right now.

Answer 7

did you try rearranging it as I suggested?

Answer 8

lol thanks, palmer. you are making my day! (:

Answer 9

a^4-81?

Answer 10

Nothing "jumped out" at me...

Answer 11

Yeah. Does that give you any ideas?

Answer 12

What's the square root of 81?

Answer 13

9, sir.

Answer 14

What's the square root of \(a^4\)?

Answer 15

...

Answer 16

i don't know...a^2?

Answer 17

\[\sqrt{a*a*a*a} = a*a\] right?

Answer 18

my cat says "if you're not getting an Oreo, you can feed me now!" brb

Answer 19

lol <3 aww you have a kitty :3

Answer 20

so the square rot of a^4 is a^2 right?
if it is, then in that case, me correcto

Answer 21

right, if \(a>0, \text{ then } \sqrt{a^4} = a^2\)

Answer 22

I see. What do I do now?

Answer 23

Well, remember the difference of squares?
\[(a^2-b^2) = (a+b)(a-b)\]

We have \[a^4-81 = (a^2)^2 - (9)^2\]

Answer 24

maybe it would be less confusing if I said the difference of squares pattern is
\[(x^2-y^2) = (x+y)(x-y)\]

Answer 25

OHHH. I see now.!!!

Answer 26

wait now i am totally confused of what to do next

Answer 27

Well, \[(x^2-y^2) = (x+y)(x-y)\]We have \[(a^4-81) = ((a^2)^2-(9)^2)\]So \(x=a^2,\,y=9\)

Answer 28

it looks so confusing, but somehow i get it

Answer 29

so write it out, what does that give us?

Answer 30

(a^2-9) (a^2 +9) ?

Answer 31

is that right?

Answer 32

Good so far. Are we done?

Answer 33

I guess I kind of let the cat out of the bag by saying "so far" :-)

Answer 34

what do you mean? yes i think so...with this problem, at least

Answer 35

Look again...

Answer 36

okay then i do not understand..

Answer 37

Okay, can you factor either of \((a^2+9)\) or \((a^2-9)\)?

Answer 38

..no...?

Answer 39

Ask grandpa :-)

Answer 40

he's asleep

Answer 41

What's 1*1?

Answer 42

What's 2*2?

Answer 43

What's 3*3?

Answer 44

1....if my memory is correct...which i am hoping it is....
2x2=4
9

Answer 45

What's a*a?

Answer 46

I might have to ask for my Oreo back :-)

Answer 47

a^2

Answer 48

My cat just said "mrrrow"...which is cat-lish for "isn't that a difference of squares?"

Answer 49

\[(a^2-9) = ((a)^2-(3)^2)\]

Answer 50

don't you have to add somehting because it is part o that formula thingy?

Answer 51

\[x = a, y = 3\]\[(x^2-y^2) = (x+y)(x-y)\]

Answer 52

yes, that so.. [(a-3) (a-3)]^2?

Answer 53

No :-(

\[x^2-y^2 = (x+y)(x-y)\]\[x=a,\,y=3\]\[x^2-y^2=(x+y)(x-y) = (a+3)(a-3)\]

Answer 54

We can't factor \((a^2+9)\) because it is a sum of squares, not a difference.

\[a^4-81 = (a^2+9)(a^2-9) = (a^2+9)(a-3)(a+3)\]

Answer 55

this problem is confusing me

Answer 56

Any time you see something squared, and a minus sign, you need to think difference of squares?

Answer 57

yes

Answer 58

Let's multiply that expression out and see what we get:

\[(a^2+9)(a-3)(a+3) = (a^2+9)(a^2 - 3a + 3a -3*3) = (a^2+9)(a^2-9) \]\[=a^2*a^2 + 9a^2 - 9a^2 -9*9\]\[ = a^4 - 81\]

Answer 59

ohhh

Answer 60

thanks!

Answer 61

We can't factor the \(a^2+9\) because if we have
\[(x+c)(x+d)\]we get those pesky +x*d + x*c and they only cancel if one of them is negative, and they have the same absolute value

Answer 62

Even \[(x+c)(x+c)\] doesn't work because\[(x+c)(x+c) = x^2 + cx + cx + c^2\]Need one of those to be \(-cx\) but if that happens then we have a \(-c^2\) at the end.

We can actually do it if we are willing to introduce imaginary numbers into the game. Have you heard of them yet?

Answer 63

uh no

Answer 64

perfect square and there's a part of it that you need to factor again