How to factor a Cubic Polynomial?

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    Cubic Polynomial
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Homework Help Overview

The discussion revolves around the factorization of a cubic polynomial, specifically f(x) = x^3 - 10x^2 + 17x + 28. Participants are exploring methods to identify potential roots that could facilitate the factorization process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants suggest guessing integer factors of the constant term to find roots, referencing the rational roots theorem. There is discussion about starting with simple integers like +1 and -1, as well as other factors of 28.

Discussion Status

The conversation is active, with participants sharing insights on identifying roots and discussing the implications of the rational roots theorem. There is a recognition that while rational roots may exist, there is no guarantee, and the exploration of potential roots is ongoing.

Contextual Notes

Participants note that the leading coefficient is 1, which influences the possible rational roots. There is an acknowledgment of the challenge in finding roots if none exist among the rational candidates.

Hilly117
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Homework Statement


Factorise:
f(x)=x^3-10x^2+17x+28
 
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First you have to guess one of the roots of the polynomial. Try substituting integer factors (divisors) of the constant term 28 in the polynomial and see which one is a root. Then you can reduce the problem to solving a quadratic equation.
 
There is a quite obvious factor to consider in such cases: +1 or -1.
 
SteamKing said:
There is a quite obvious factor to consider in such cases: +1 or -1.
:-p

Hilly172, both Hilbert2 and SteamKing are using the "rational roots theorem":
if rational number m/n satisfies the polynomial equation a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0, then the numerator, m, must evenly divide the "constant term", a_0, and the denominator, n, must evenly divide the "leading coefficient", a_n". Here, the leading coefficient is 1 and the only positive integer that divides that is 1 so any rational solution must have denominator 1- that is, must be an integer. And that integer must evenly divide, so must be a factor of, the constant term, 28. And the simplest such integers to start with are 1 and -1. The other factors of 28 are, of course, 2, -2, 4, -4, 7, -7, 14, and -14.

Of course, there is no guarentee that there is a rational root. But if there is not, the solution to the equation is going to be very difficult so it is worth trying. (And, as you have probably guessed from what Hilbert2 and SteamKing said, here, there is a very simple solution.)
 
Last edited by a moderator:
I didn't consciously set out use the RRT, I just scanned the polynomial to see if +1 or -1 would make it zero. For a cubic, you just need one root obtained by guessing or by plotting, and then Bob's your uncle.
 

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