MHB Real Zeros of a Polynomial Function

[Taryn1]

So I'm supposed to find all the real zeros of this polynomial function:

$\int$ $\left(x\right)$ = $x^3$ + 3$x^2$ - 4$x$ - 12

Usually, to find the zeros, I would use the quadratic function

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

But what do I do with the 3 at the beginning of the function? I should probably already know this from algebra, but I can't factor any numbers out and frankly I'm not sure what to do. I tried working it using the quadratic function with 3 = a, -4 = b, and -12 - c, but I got messy answers like this:

$\frac{4 \pm 4\sqrt{10}}{6}$

and the answers in the back of the book are $\pm$2, -3. I don't know how to get there, though. Help!

- - - Updated - - -

P.S. Special thanks to Ackbach for helping me get the mathy stuff to enter right (Clapping)

 

[Greg]

Hint: $$f(-2)=0$$

 

[Taryn1]

(Wondering) How do you get that from the equation? By guessing and plugging in -2 and seeing if it equals zero?

...because I have the answers in the back of the book, but I don't know how to technically get them!

 

[Greg]

Yes; it's often called solving "by inspection".

You may now use polynomial long division to find the other two roots (factor the quadratic resulting from this division).

 

[Taryn1]

Oooohhhhh, right. Polynomial long division. I knew it was something I should have remembered from algebra!

Thanks :)

 

[MarkFL]

Using the rational roots theorem, we know that the rational roots, if there are any, must come from the list:

$$\pm1,\,\pm2,\,\pm3,\,\pm4,\,\pm6$$

If we define $P(x)=x^3+3x^2-4x-12$, then proceeding systematically, we find:

$$P(1)=-12$$

$$P(-1)=-6$$

$$P(2)=0$$

Okay, now since we have a cubic, division by this zero will result in a quadratic, which we can hopefully factor by hand. Using synthetic division, we find:

$$\begin{array}{c|rr}& 1 & 3 & -4 & -12 \\ 2 & & 2 & 10 & 12 \\ \hline & 1 & 5 & 6 & 0 \end{array}$$

So, we now know:

$$P(x)=(x-2)\left(x^2+5x+6\right)=(x-2)(x+2)(x+3)$$

Hence, by the zero-factor property we obtain the 3 real roots for $P$ of:

$$x\in\{-3,\pm2\}$$

 

[Ackbach]

So I'm supposed to find all the real zeros of this polynomial function:

$\int$ $\left(x\right)$ = $x^3$ + 3$x^2$ - 4$x$ - 12

You can actually do better with the following code:

$f(x)=x^3+3x^2-4x-12$

producing

$f(x)=x^3+3x^2-4x-12$.

You don't have to come in and out of math mode while you're in the middle of an expression, unless there's actual text you need to display.

Cheers!

 

[mathbalarka]

You can always factorize.

$x^3 + 3x^2 - 4x - 12 = x^3 - 4x + 3x^2 - 12 = x(x^2 - 4) + 3(x^2 - 4) = (x^2 - 4)(x + 3) = (x + 2)(x+3)(x - 2)$

Hence, the roots are $x = 2, -2, -3$.

 

[Deveno]

Here is something well worth remembering:

If $a$ is a root of $p(x)$, then $(x - a)$ is a FACTOR of $p(x)$, and vice versa.

The proof of this is straightforward:

Divide $p(x)$ by $(x - a)$ to get:

$p(x) = q(x)(x - a) + r$, where $\text{deg}(r) < \text{deg}((x-a))$, or $r = 0$.

Since the degree of $x - a$ is 1, this means $r$ will be a NUMBER (constant polynomial).

Which number is it? Letting $x = a$, we get:

$p(a) = q(a)(a- a) + r = q(a)0 + r = 0 + r = r$.

So $r = p(a)$. So we always have:

$p(x) = q(x)(x - a) + p(a)$.

If $p(a) = 0$, then $p(x) = q(x)(x - a)$, and we see that $(x - a)$ is indeed a factor (and we have shown the factorization, produced by division).

On the other hand, if $p(x) = q(x)(x - a)$, for some polynomial $q(x)$ then:

$p(a) = q(a)(a - a) = q(a)0 = 0$, and $a$ is a root of $p$.

Thus there is a $1-1$ correspondence between roots of $p$ (these are numbers) and linear factors $(x - a)$ (these are polynomials).

A word about the "rational roots test": it is a fact (a deep one, which I shall not prove here) that if a polynomial with INTEGER coefficients factors into two (or more) polynomials with RATIONAL coefficients, then we can ALSO find a factorization with INTEGER coefficients.

In your case, since you have a cubic, the factorizations (if they exist) must be:

$x^3 + 3x^2 - 4x - 12 = (ax + b)(cx + d)(mx + n)$ or:

$x^3 + 3x^2 - 4x - 12 = (ax + b)(sx^2 + tx + u)$

where $a,b,c,d,m,n,s,t,u$ are INTEGERS.

In case 1, we have $acm = 1$, in case 2, we have $as = 1$, and since these are all INTEGERS, $a = \pm 1$. By appropriately assigning signs, we can force $a = 1$. So if we have a RATIONAL factorization (it might be we don't), we have $(x \pm b)$ as a factor.

Now either $bdn = -12$, or $bu = -12$, either way, we see that $b$ is a factor of $-12$, so the only possible rational roots to check turn out to be integers(!) which are factors of $-12$, and there are only so many of these:

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$ or $\pm 12$.

Whittling down the infinity of real numbers to only 12 possibilities is a minor triumph, of sorts, and $p(a)$ for these $12$ integers is easily checked.