Question

Fan and Medal for one question :)

Answer 1

9.)Factor 2g^3-g^2-8g+4

Answer 2

Only one?

Answer 3

yep

Answer 4

There are four terms, separate and group 2 terms..

Answer 5

Ok?

Answer 6

\[2g^3-g^2-8g+4 \implies \color{green}{(2g^3 - g^2)} + \color{red}{(-8g + 4)}\]

Answer 7

Now, solve for first green terms.

Answer 8

g2(2g−1)

Answer 9

You can see easily, that first term is having 3 g's and second is having 2 g's, so you can take 2 g's as common from it like this:
\[\color{green}{g^2(2g - 1)}\]

Answer 10

Up to here, all good?

Answer 11

yep

Answer 12

-4(2g-1) for red?

Answer 13

Now, work on the red part, you will see that, you can take \(4\) as common from it, also, you can take \(-\) sign common..

Answer 14

yep that is excellent, you reduced my work.. :P

Answer 15

lol

Answer 16

So, finally, you are having:
\[\color{green}{g^2(2g - 1) } \color{red}{-4(2g-1)}\]

Answer 17

Now, you will see that there is common term in both green and red and that is : \((2g-1)\)

Answer 18

yes

Answer 19

So, take \((2g-1)\) common from both green and red part:
\[\color{blue}{(2g-1)} \cdot ( \; \color{green}{g^2} \color{red}{-4})\]

Answer 20

All good up to here?

Answer 21

Ok. yep

Answer 22

Is it finished?
No, not yet, you have to wait more, you can use one formula here:
\[a^2 - b^2 = (a+b)(a-b)\]

Answer 23

\(a = g\) and \(b = 2\)

Answer 24

So:
\[g^2 - 2^2 = (g+2)(g-2)\]

Answer 25

So, finally you are having three factored terms:
\[\implies \color{blue}{(g+2)} \cdot \color{red}{(g-2)} \cdot \color{green}{(2g-1)}\]

Answer 26

Any doubt?

Answer 27

i am sorry my internet cut out. i got the right answer thnak you soo much:)

Answer 28

\(\dagger\)