1
$\begingroup$

I have this:

$f(x) = 0$

where

$f(x) := \cfrac{3x^2 - 5x + 2}{x + 2}$

How do I solve that?

Do I multiply by $(x + 2)$ and solve $3x^2 - 5x + 2=0$ or solve $3x^3 + x^2 - 8x + 4=0$ with Horner method?

  • 0
    Yes, you solve for the roots of the numerator of your rational function. You do not solve for the roots of $3x^3 + x^2 - 8x + 4$ since that is more complicated, and one of the roots of that polynomial is not a root of your original rational function.2012-08-09
  • 0
    In plain words, do I solve it with Horner? so it's like a/b = 0 => a*b = 0?2012-08-09
  • 1
    The denominator doesn't enter into finding the roots, so just take the polynomial on top and solve for its roots.2012-08-09
  • 0
    so I solve 3x^2 - 5x + 2 = 0?2012-08-09
  • 0
    Yes.$\phantom{}$2012-08-09
  • 0
    It seems to me unclear what your function is - is it $3x^2-5x+\frac 2 {x+2}$ or $\frac {3x^2-5x+2}{x+2}$ or $3x^2-5x+\frac 2 x +2 $2012-08-09
  • 0
    @Mark its the second2012-08-09
  • 0
    @Cameron thanks for the edit :)2012-08-09
  • 0
    Actually, the denominator *does* enter into finding the roots: After you've found the roots of the numerator, you have to check that they are *not* roots of the denominator.2013-07-24

1 Answers 1

8

In plain words, if $a/b$ is defined and equal to $0$, then $a=0$. Of course, this does imply that $a\cdot b=0,$ but that isn't relevant. You need only determine the solutions of $3x^2-5x+2=0$.

  • 0
    I solved it as a*b = 0 and got 1 and -2 as roots2012-08-09
  • 0
    @Nick, $-2$ is not a valid solution. Try substituting it into your original expression.2012-08-09
  • 0
    There should be another solution to that quadratic. Also, what is $f(-2)$? It's worth noting that $a\cdot b=0$ **does not** imply that $a/b=0$. It implies that $a/b=0$ or is undefined.2012-08-09
  • 0
    With Horner ( a* b = 0) I got 1 and -2 but by solving the polynomial at the top I got 1 and 2/3 as roots2012-08-09
  • 0
    My first thought was to solve it with a*b = 0 but my brother insisted to only find the roots of the top polynomial2012-08-09
  • 0
    $1$ and $2/3$ are the appropriate roots. One point of clarification, though. Consider the function $$g(x):=\frac{3x^2-5x+2}{x-1}.$$ Here, the roots of the numerator are the same. However, $1$ is **not** a zero of $g(x)$, since $g(x)$ is undefined at $x=1$.2012-08-09
  • 2
    @Nick: Forget Horner; Horner's method has absolutely nothing to do with this problem. To make it extremely clear: $$\frac{a}{b} = 0 \Leftrightarrow a=0$$ Now $\frac{a}{b} = 0 \Rightarrow ab=0$, but $ab=0 \not \Rightarrow \frac{a}{b} = 0$. In this problem we *only* care about when $a=0$.2012-08-09
  • 3
    That is, $\frac{a}{b}=0$ if and only if $a=0$ and $b\neq 0$.2012-08-09
  • 0
    Okay, thanks for your help everyone. Much appreciated.2012-08-09