2
$\begingroup$

As I'v learned about polynomials, I run into this quadrinomial:

$$x^3+300x^2+30000x-953125 = 0$$

I've been studied how to factor this quadrinomial but didn't quite understand how it's done, here is how I've studied(I probably didn't get the following steps wrong):

I should ask myself how big x has to be for the linear term and the constant to add to 0.

Divide the constant by the coefficient of the linear term, its around 30, now how would I know if 30 is a reasonable approximation for x?

I should check the result when x = 30 for the quadratic and cubic terms and if each of them is smaller than the constant then it's a reasonable approximation.

As I checked with other polynomials dividing the constant with the linear term always produce a number that is pretty close to the x.

I know I'm probably misunderstood most of the things, but if anyone can help me figure out why and how I will be very thankful.

Edit:

I'm sorry for not being clear about my question, I try to understand why after dividing the constant by the linear and getting something around 30, why It's considered as reasonable approximation for the factor when both of the following came up true: $$30^3 < 30*30,000$$ and $$300*30^2 < 30*30,000$$ It's something I've learned and couldn't understand why it's true to do so.

$\endgroup$
0

2 Answers 2

1
$\begingroup$

You should use the fact that if your polinomial $$ f = x^3+300 x^2+30000 x-953125 $$ has an integer root $\alpha$ (so that $x - \alpha$ is a divisor of $f$), then $\alpha$ divides the constant term $953125 = 5^{6} \cdot 61$. (Also, a rational root of $f$ has to be an integer, as $f$ is monic.)

Try the divisors of $5^{6} \cdot 61$, starting with the small positive ones, and you will soon find a root$\dots$

$\endgroup$
1
  • $\begingroup$ You're welcome! $\endgroup$ Mar 18, 2015 at 21:51
1
$\begingroup$

Hint. Since you are looking for a zero which is a divisor of $953125 =5^{6} \cdot 61$ and which is near $30$, then you may supect it to be equal to $\color{red}{25}$, this may lead you to write $$ \begin{align} & x^3+300x^2+30000x-953125\\\\ =\:&\left(x^3+3\color{red}{25}x^2+3\color{red}{8125}x\right)-\left(\color{red}{25}x^2+\color{red}{8125}x+953125\right)\\\\ =\:&\color{red}{x}\left(x^2+325x+38125\right)-\color{red}{25}\times\left(x^2+325x+38125\right)\\\\ =\:&(\color{red}{x-25})\left(x^2+325x+38125\right). \end{align} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .