Question

Explain, in complete sentences, how you would factor 12x^3– 82x^2– 14x

Answer 1

First, factor out the GCF. What do you think that is? \[12x^3 - 82x^2 - 14x\]

Answer 2

2x

Answer 3

Ok. \[12x^3 - 82x^2 - 14x \implies 2x(6x^2 - 41x - 7)\]Now, do you remember how to factor by grouping?

Answer 4

umm..not exactly

Answer 5

Ok. multiply a and c of \(ax^2 + bx + c\) and find factors of that that add up to -41

Answer 6

well a and c make -42

Answer 7

Ok. Now find factors of -42 that add up to -41.

Answer 8

so -42 and +1

Answer 9

Ok. Split up the middle term using those factors.
\[2x(6x^2 - 42x + x - 7) \implies 2x((6x^2 - 42x) + (x - 7))\]Can you factor out the GCF of both groupings? as in 6x^2 - 42x and x - 7

Answer 10

there is no further GCF...i think

Answer 11

is there?

Answer 12

How about the first group? GCF of 6x^2 - 42x

Answer 13

oh 6x(x-7)

Answer 14

but theres nothing for x-7

Answer 15

wait so its 2[6x(x-7)^2)]

Answer 16

??

Answer 17

i dont think thats completely right :\

Answer 18

Alright. \[2x((6x^2 - 42x) + (x - 7)) \implies 2x(6x(x - 7) + (x - 7))\]Now, using the distributive property, bring them together. \[2x(6x(x - 7) + (x - 7)) \implies 2x(6x + 1)(x - 7)\]Did you follow?

Answer 19

yea i do but shudnt this polynomial only end up with 3 factored parts?

Answer 20

or wait does x-7 count as just one?

Answer 21

being 2x, 6x, and x-7?

Answer 22

What do you mean? I used the distributive property for this. that's why it turned to \((x - 7) \space and \space not \space(x - 7)^2\)
Also, the beginning degree was 3, so there are three solutions or factors.

Answer 23

yea the beginning degree was 3 meaning there are 3 solutions/factors...but this one has four: 2x, 6x, x-7, and x-7

Answer 24

oh wait....its 2x, 6x, and 2x-14

Answer 25

right?

Answer 26

No. I used the distributive property. Well, ok.
Think of it like this. Let's say that \(x - 7 = y\)
Now let's substitute stuff now.
\[2x(6x(x - 7) + (x - 7)) \implies 2xy + y \implies y(2x + 1) \implies (x - 7)(2x + 1)\]

Answer 27

okay so can you just tell me what the 3 FINAL solutions/factors are? :)

Answer 28

Alright. The final factors are 2x, x - 7, and 2x + 1

Answer 29

where did you get 2x+1 :'(

Answer 30

Oops. I meant 6x + 1

Answer 31

Writing a book there? lol

Answer 32

okay so heres what my answer would be:

pull out the GCF:
2x(6x^2-41x-7)
factor out the parenthesis:
2x[(6x^2-42x)+(x-7)]
that can be further factored to:
2x[6x(x-7)+(x-7)]
solutions:
2x, 6x+1, x-7


^^good enough? :P

Answer 33

well??

Answer 34

Close. This is what I'd say.

pull out the GCF:
2x(6x^2-41x-7)
split the middle term and group the terms
2x[(6x^2-42x)+(x-7)]
factor out the GCF within the groups
2x[6x(x-7)+(x-7)]
distributive property
2x[(6x+1)(x-7)]
final factors
2x, 6x+1, x-7

Answer 35

even better :D ...thanks! :)

Answer 36

np :)

Answer 37

:)