Question

Multiply PLEASE HELP

Answer 1

Use this property.

\(\Large \color{Black}{\Rightarrow {a \over b} \times {x \over y} = {ax \over by} }\)

Answer 2

The denominator will remain the same. Multiply numerators.

Answer 3

ok just amin im working on it

Answer 4

\(\Large \color{Black}{\Rightarrow {(k + 3)(12k^2 + 2k -4) \over 4k - 2} }\)

Answer 5

im not sure is that the answer because its not in the options

Answer 6

Simplify the numerator.

Answer 7

\(\Large \color{Black}{\Rightarrow (k + 3)(12k^2 + 2k - 4) = k(12k^2 + 2k - 4) + 3(12k^2 + 2k -4) }\)

Answer 8

Oops

Answer 9

\(\Large \color{Black}{\Rightarrow k(12k^2 + 2k - 4) + 3(12k^2 + 2k - 4) }\)

Answer 10

3 (k + 3) 2k-1/2

Answer 11

What?

Answer 12

lol just a min lemme re type that

Answer 13

i think this is the answer...im not sure

Answer 14

\(\Large \color{Black}{\Rightarrow 12k^3 + 2k^2 - 4k + 36k^2 + 6k - 12 }\)

Answer 15

Be patient.

Answer 16

ok...

Answer 17

\(\Large \color{Black}{\Rightarrow 12k^3 + 38k^2 +2k - 12 }\)
\(\Large \color{Black}{\Rightarrow 2k(6k^2 + 38k + 1) - 12 }\)

Answer 18

this is option b

Answer 19

c.(k + 3)(3k + 2)
d.(3k + 1)(2k + 3)

Answer 20

basically im still cnnfused i dont know how ot get a solid answer

Answer 21

You may use a calculator, or wolfram.

Answer 22

hold on

Answer 23

im so confused eight now

Answer 24

right*

Answer 25

Lol no need.

Answer 26

Just factor the numerator.

Answer 27

Factorise the numerator and the denominator. Then cancel out the common factors.

Answer 28

A better way is to divide both numerator and denominator by 2.

Answer 29

i think i got the answer thanks guys

Answer 30

Hmm you bet :)