Question

8p^5-200p^3+27p^2-675

I need help factoring... Steps, please!

Answer 1

\[8p^5-200p^3+27p^2-675\]

Answer 2

Factor by grouping will probably work.

Answer 3

Even with 5th degree polynomials?

Answer 4

Any time you have a four term polynomial, it's worth a shot.

Answer 5

Ok, thanks! I'll try it.

Answer 6

Me too. Let me know what you get and I'll check it.

Answer 7

All right.

Answer 8

(Ooh, this one falls apart quite nicely..)

Answer 9

Ok, I ended up with \[(p-5)(p+5)(2p+3)(4p^2-12p+9)\]

Answer 10

mm, I'm getting (4p^2 - 6p +9) for that quadratic.

Answer 11

I just checked the back of the book though, and the answer is the same thing except that the last bit is \[(4p^2-6p+9)\], but I have no idea what I did wrong.

Answer 12

OH. i see what I did!!! yay, I got it right!
I just multiplied 4 and 3 instead of 2 and 3.

Answer 13

Thanks! A lot!

Answer 14

Bonus question: Can you use those factors to find the roots of the polynomial?

Answer 15

uhhhh... yes?


Here goes.

Lets see...

Answer 16

>.< my computer froze, sorry!!! gahhh... ok, so is this right:

(p -5)*(p +5)*(2*p +3)*(4*p^2 -12*p +9) = 0 =>
=> 8*p^5 -12*p^4 -218*p^3 +327*p^2 +450*p -675 = 0
=> 8*(p +5)*(p +1.5)*(p -1.5)*(p -1.5)*(p -5) = 0

-5, -1.5, 1.5, 1.5, 5

Answer 17

I lost connection last night too. The three real roots, -5, -1.5, and 5 are right, but remember: the quadratic is 4p^2 -6p +9, not 4p^2 -12p +9. If you try quadratic formula on that, you'll get two non-real solutions.

Looks like you know what you're doing, though, just be a little more careful writing down the correct numbers.

Nicely done!