Explain, in complete sentences, how you would factor 12x^3– 82x^2– 14x
First, factor out the GCF. What do you think that is? \[12x^3 - 82x^2 - 14x\]
2x
Ok. \[12x^3 - 82x^2 - 14x \implies 2x(6x^2 - 41x - 7)\]Now, do you remember how to factor by grouping?
umm..not exactly
Ok. multiply a and c of \(ax^2 + bx + c\) and find factors of that that add up to -41
well a and c make -42
Ok. Now find factors of -42 that add up to -41.
so -42 and +1
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
there is no further GCF...i think
is there?
How about the first group? GCF of 6x^2 - 42x
oh 6x(x-7)
but theres nothing for x-7
wait so its 2[6x(x-7)^2)]
??
i dont think thats completely right :\
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?
yea i do but shudnt this polynomial only end up with 3 factored parts?
or wait does x-7 count as just one?
being 2x, 6x, and x-7?
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.
yea the beginning degree was 3 meaning there are 3 solutions/factors...but this one has four: 2x, 6x, x-7, and x-7
oh wait....its 2x, 6x, and 2x-14
right?
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. \[2x(6x(x - 7) + (x - 7)) \implies 2xy + y \implies y(2x + 1) \implies (x - 7)(2x + 1)\]
okay so can you just tell me what the 3 FINAL solutions/factors are? :)
Alright. The final factors are 2x, x - 7, and 2x + 1
where did you get 2x+1 :'(
Oops. I meant 6x + 1
Writing a book there? lol
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
well??
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
even better :D ...thanks! :)
np :)
:)
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