I do not understand how we know a cubic polynomial such as

z^3 -3z^2 + 4z - 2 = 0

can be factored as

(z - 1)(z^2 - 2z +2) = 0

I can, of course, work it backwards but is this actually some kind of identity? and if so what is its derivation?

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- Nov 14th 2012, 06:44 AMalyosha2Factoring cubic polynomials
I do not understand how we know a cubic polynomial such as

z^3 -3z^2 + 4z - 2 = 0

can be factored as

(z - 1)(z^2 - 2z +2) = 0

I can, of course, work it backwards but is this actually some kind of identity? and if so what is its derivation? - Nov 14th 2012, 07:25 AMBobPRe: Factoring cubic polynomials
At this sort of level, it's pretty much a case of trial and error.

Substitute values for z and hope that one of them gets you a zero value for the polynomial.

Start off by trying z = 0,1,2,-1,-2,3 etc. For this particular example you get a hit with z = 1 in which case z - 1 is a factor and you can now divide out to determine the quadratic.

Notice though that if for two of your trial values the polynomial changes sign then there has to be a zero between those two values. So for your example z = 0 and z = 2 gets you polynomial values of -2 and +2 in which case there has to be a value of z between 0 and 2 making the poynomial zero. (In general this need not be an integer value.)

If you get no joy from this method you might consider sketching a graph or graphs (usually graphs is better).

For this example rewrite the equation as $\displaystyle z^{3}-3z^{2}=2-4z$ and sketch the graphs of $\displaystyle y = z^{3}-3z^{2}$ and $\displaystyle y = 2 -4z.$ The values of z at which the graphs intersect will be the roots of the equation.

For this example this will tell you that there is a root somewhere between 1/2 and 3. - Nov 14th 2012, 09:54 AMSorobanRe: Factoring cubic polynomials
Hello, alyosha2!

Quote:

I do not understand how we know a cubic polynomial such as: .$\displaystyle x^3 -3x^2 + 4x - 2 \:=\: 0$

can be factored as: .$\displaystyle (x - 1)(x^2 - 2z +2) \:=\:0$

We are given a polynomial equation: .$\displaystyle P(x) \:=\:0$

We are expected to know two theorems.

[1] If $\displaystyle P(a) = 0$, then $\displaystyle x = a$ is a root of the equation

. . .and $\displaystyle x-a$ is a factor of $\displaystyle P(x).$

[2] If $\displaystyle P(x)$ has a rational root, it is of the form: $\displaystyle x \,=\,\frac{n}{d}$

. . .where $\displaystyle n$ is a factor of the constant term

. . .and $\displaystyle d$ is a factor of the leading coefficient.

Our polynomial is: .$\displaystyle x^3 - 3x^2 + 4x - 2$

Its constant term is 2; its factors are: $\displaystyle \pm1,\:\pm2$

Its leading coefficient is 1; its factor are: $\displaystyle \pm1$

Hence, the only possible rational roots are: .$\displaystyle \pm1,\:\pm2$

So, we try each of them until one of them makes $\displaystyle P(x)$ equal to zero.

We find $\displaystyle x = 1$ is a zero of the polynomial.

. . Hence, $\displaystyle (x-1)$ is a factor.

Using long division, we have: .$\displaystyle x^3 - 3x^2 + 4x - 23 \:=\:(x-1)(x^2-2x+2)$