In my Algebra II class we are learning how to find the zeros of a function, but I find the process very confusing despite the many efforts of my algebra teacher to explain them to me. I understand that there are different methods to do these, and I find quartic functions very easy to manage because of how you can replace $x^4$ with $a^2$ in order to use the quadratic formula, but how would you do more complicated ones such as cubics? I am struggling with my homework assignment in general, but here is one that I find particularly troubling: $f(x) = x^3 - 2x^2 - 3x + 6$. Could someone explain to me how to do this one? I am fairly certain that this one has imaginary answers, but that's only a guess. Please don't just post a link to another website, as I have looked at every imaginable page on Google but to no avail.
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$\begingroup$Notice that factor-by-grouping gives $$x^3 - 2x^2 - 3x + 6 = x^2(x - 2) - 3(x - 2) = (x^2 - 3)(x - 2)$$
Then, find the zeroes, which is easy. That is, find the values of $x$ such that
$$x^2 - 3 = 0\text{ and }x - 2 = 0$$
This should be simple for you to work out. ;)
Another way to determine the zeroes is to use Rational Root Theorem and synthetic division, which requires more work than factorization. But it's quite useful to determine the rational zeroes (only if they exist in the polynomial!).
For the given problem, the first coefficient is $1$ and the last coefficient is $6$. By Rational Root Theorem, we have
$$\left\{\pm \dfrac{\text{factors of last term}}{\text{factors of first term}}\right\} = \left\{\pm \dfrac{1,2,3,6}{1} \right\} = \{\pm 1,\pm 2, \pm 3,\pm 6\}$$
Select one of the rational roots from the set above and substitute for $f(x)$. If $f(a) = 0$ for $a \in \{\pm 1, \pm 2, \pm 3, \pm 6\}$, then $a$ is the root, so we can use synthetic division. Otherwise, select another rational root and try again. Otherwise, if all rational roots fail, try approximation.
For answers, scroll over this shaded box.
$\endgroup$ $\begingroup$The zeroes are $x = \pm\sqrt{3}$ and $x = 2$
Hint: what is the value of $f(2)$?
$\endgroup$ 3 $\begingroup$General cubics are most easily solved by guesswork (perhaps aided by Descartes's rule of signs or other bounds). But some cubics are nicer, including your example $f(x)=x^3−2x^2−3x+6$. Note that the four terms come in two pairs, $x^3−2x^2$ and $−3x+6$, that match up very nicely: the first is $-x^2/3$ times the second. Thus, you get $f(x)=(-x^2/3+1)(-3x+6)$, which you can (simplify if you want factors and) use to find zeros.
$\endgroup$ $\begingroup$Generally, you aren't going to be able to factorize every polynomial you see. Furthermore, fifth-degree polynomials and higher don't even always have roots that can be written with radicals.
However, in this case, you can factorize the expression:
Note that $x^3-2x^2-3x+6$ = $x^2(x-2) - 3 (x-2)$ = $(x^2-3)(x-2)$ I assume you can proceed.
If the polynomial doesn't have any clear factorization like this, nor is a power of a binomal, you're generally out of luck. There are cubic and quartic formulas, but they're far from practical. The rational root theorem can help, but as you move into higher levels of mathematics, you will be able to find approximations of the zeroes using various methods.
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