Question

6x^2(4x^2+3)

Answer 1

Your try to solve for x?

Answer 2

hey after this, can u go back to ur previous question, ive replied to ur question

Answer 3

@mparm

Answer 4

It says to factor each polynomial

Answer 5

First set it equal to 0 :
\[6x^2(4x^2+3) = 0\]
Divide both sides by 6:
\[x^2(4x^2+3) = 0\]
Split into two equations:
\[x^2 = 0 or 4x^2 + 3 = 0\]
x = 0 or 4x^2 = -3

Divide both sides by 4:

x = 0 or x^2 = -3/4

Answer 6

So would I just answer it like that on my homework? and could u help me with some other questions please?

Answer 7

Because x^2 = -3/4 has no solution since for all x on the real line, x^2 >= 0 and -3/4 < 0:

x = 0 is the anser

Answer 8

Well, are you looking for Real roots or Complex roots?

Answer 9

i dont know it just tells me to factor each polynomial

Answer 10

Your sure the problem is written correctly?

Answer 11

yes

Answer 12

oh it says factor each product

Answer 13

i ment simplify each product it says

Answer 14

can u help me with 16b^4+8b^2+20b it says factor each polnomial

Answer 15

For the second one you could factor out 4b

resulting in:
\[4b(4b^3 + 2b + 5)\]

Answer 16

ok so that be the answer?

Answer 17

I can't see any other way to simplify it, but I'm not the best at this sort of thing

Answer 18

I have to go work on some school now though :/

Answer 19

ok thnx

Answer 20

ohh i still need help im soo behind

Answer 21

does this look right (x+2)(x+9)
X^2+11x+18
¼(2x+11)^2- 49/4
X(x+11)+18
?