Question

Simplify completely: the quantity 6 times x to the 4th power plus 9 times x to the 2nd power plus 12 times x all over 3 times

Answer 1

a. 2x3 + 9x2 + 12
b. 2x2 + 3x + 4
c. 2x5 + 3x3 + 4x2
d. 2x3 + 3x + 4

Answer 2

Can you write out your function first?

Answer 3

Here it is

Answer 4

Since x is common to all terms in both numerator and denominator, and 6,9, and 12 are a multiple of 3, reduce each term in the numerator by 3x

Answer 5

so \[\frac{6x^4}{3x} ~,~ \frac{9x^2}{3x} ~,~ \frac{12x}{3x}\]

Answer 6

I am lost

Answer 7

what is 6/3?

Answer 8

2

Answer 9

and \(\large x^{4-1}\)?

Answer 10

x^3

Answer 11

That takes care of the first part of the problem. You get \(2x^3\).

Answer 12

Now what is 9/3 and \(\large x^{2-1}\)?

Answer 13

3 * x^1

Answer 14

and so the second part gives you \(3x\).

Answer 15

Now for the final part:

12/3 and \(x^{1-1}\)

Answer 16

4x

Answer 17

\[x^{1-1} =x^0 =1\] That's just an identity you learn and get comfortable with.

Answer 18

So therefore it'd just be 4.

Answer 19

Now we just combine all our results together to get: \(2x^3+3x+4\)

Answer 20

That isn't one of the options

Answer 21

Nevermind

Answer 22

:)