Question

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

Answer 1

Do you know group factoring?

Answer 2

No

Answer 3

Group factoring is like:

\[x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x+1)(x^2+1)\]

Answer 4

I don't really remember learning about that, but I think I understand a little bit

Answer 5

so you need to group factor the numerator

Answer 6

oh, wait. It can't be group factored >.<

Answer 7

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
\[\pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 3 } ~\]

Answer 8

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.

Answer 9

can someone explain? ;-;

this is hard to show online compared to in-person with a piece of paper ;-;

Answer 10

someone else*

Answer 11

long division

Answer 12

ik, but latex isn't that easy to use to show that ;-;

and the draw tool isn't that easy either ;-;

Answer 13

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

Answer 14

haha true that !

Answer 15

divide first terms \[\huge\rm \frac{ 3x^3 }{ 3x } = ???\]
then multiply the divisor with that