Question

-12a(to the 4th) + 9a(to the 3rd) -12a(to the 2nd) = ??

Answer 1

First thing you should do is factor out the equation
-12a^4+9a^3-12a^2=
-3a is the number you can use to factor it out.
and you get
-3a(4a^3+3a^2-4a)

Answer 2

pull out another a

Answer 3

OH sorry you can pull out -3a^2

Answer 4

-3a^2 ( 4a^2 -3a +4)

Answer 5

Now you can factor what's in the ( )

Answer 6

So All together it goes like this (correct me if i am wrong)..
-12a^4+9a^3-12a^2=
-3a is the number you can use to factor it out.
and you get
-3a^2 ( 4a^2 -3a +4)
and then you factor the inside which you get (2a+1)(2a-4)
Leading to -3a^2(2a+1)(2a-4) which is the answer

Answer 7

I'm getting imaginary numbers. (the square root of a negative). Is that something you've dealt with before?

Answer 8

it goes like this: -3a^2(4a^2-3a+4)

Answer 9

yes

Answer 10

You have to factor the inside (4a^2-3a+4)

Answer 11

Then I used the quadratic formula to factor the inside

Answer 12

what would be the result ? just multiplying and adding to simplify

Answer 13

Hm Ceriona did i make a mistake?

Answer 14

Oh wait (4a^2-3a+4) the +4 should be -4

Answer 15

so it is (4a^2-3a-4) which leads to the answer i have put

Answer 16

a = (3+ √55 i)/8 and
a = (3- √55 i)/8 as imaginary roots
and a=0 as a double root

Answer 17

Here is what i did i dont think quadratic formula is neccesary
-12a^4 +9a^3 -12a^2
-3a^2(4a^2-3a-4)
-3a^2(2a-4)(2a-1) = Answer

Answer 18

lol thans but u are wrong

Answer 19

if there is an imaginary number ( the dicriminant is a negative number) it means that it does not pass the x axis, meaning, there is only one real root in the equation.

Answer 20

There is no imaginary number though. Answer should be -3a^2(2a-4)(2a-1) if you factor it out. I dont know why she said i am wrong unless she shows me what i did wrong

Answer 21

(2a-4)(2a-1)=4a^2-10a+4
not 4a^2-3a+4

Answer 22

Aha thank you very much :D