## anonymous one year ago What is the quotient: (3x^3 + 14x^2 + 13x - 6) / (3x - 1) ?

1. AbdullahM

Do you know group factoring?

2. anonymous

No

3. AbdullahM

Group factoring is like: $x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x+1)(x^2+1)$

4. anonymous

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

5. AbdullahM

so you need to group factor the numerator

6. AbdullahM

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

7. AbdullahM

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 $\pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 3 } ~$

8. AbdullahM

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.

9. AbdullahM

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

10. AbdullahM

someone else*

11. Nnesha

long division

12. AbdullahM

ik, but latex isn't that easy to use to show that ;-; and the draw tool isn't that easy either ;-;

13. Nnesha

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

14. Nnesha

haha true that !

15. Nnesha

|dw:1442023891316:dw| divide first terms $\huge\rm \frac{ 3x^3 }{ 3x } = ???$ then multiply the divisor with that

16. Nnesha

^ divide first term of polynomial by first term of the divisor

17. Nnesha

wait for ookay :=)

18. AbdullahM

thanks :)

19. Nnesha

yw :=)