What is the quotient: (3x^3 + 14x^2 + 13x - 6) / (3x - 1) ?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

What is the quotient: (3x^3 + 14x^2 + 13x - 6) / (3x - 1) ?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Do you know group factoring?
No
Group factoring is like: \[x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x+1)(x^2+1)\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I don't really remember learning about that, but I think I understand a little bit
so you need to group factor the numerator
oh, wait. It can't be group factored >.<
The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction p/q, where p is a factor of the trailing constant and q is a factor of the leading coefficient. and so using that, these could be possible roots \[\pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 3 } ~\]
solve 3x-1 = 0 x = 1/3 and since that is one of the possible roots, we can use that since it might be divisible by it.
can someone explain? ;-; this is hard to show online compared to in-person with a piece of paper ;-;
someone else*
long division
ik, but latex isn't that easy to use to show that ;-; and the draw tool isn't that easy either ;-;
for long division 5 steps) \(\huge\color{green}{{1:}}\) Divide the first terms \(\huge\color{green}{{2:}}\) multiply(distribute) \(\huge\color{green}{{3:}}\) subtract all terms \(\huge\color{green}{{4:}}\) carry down \(\huge\color{green}{{5:}}\) repeat
haha true that !
|dw:1442023891316:dw| divide first terms \[\huge\rm \frac{ 3x^3 }{ 3x } = ???\] then multiply the divisor with that
^ divide first term of polynomial by first term of the divisor
wait for `ookay` :=)
thanks :)
yw :=)

Not the answer you are looking for?

Search for more explanations.

Ask your own question