Solve Simple Polynomial: Find a if fx=2x^3-ax^2+2x-7

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Homework Help Overview

The discussion revolves around finding the value of 'a' in the polynomial function fx = 2x^3 - ax^2 + 2x - 7, given that it has a repeated factor. The subject area includes polynomial functions and their properties, particularly focusing on roots and factors.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various methods to determine 'a', including trial and error, the Rational Roots Theorem, and synthetic division. Questions arise regarding the nature of repeated factors and the implications of having double or triple roots.

Discussion Status

The discussion is active, with multiple approaches being suggested. Some participants express concern about the efficiency of trial and error methods, while others propose using polynomial identities and derivatives to find the roots. There is no explicit consensus on the best method yet.

Contextual Notes

Participants note that the problem is part of a Further Math syllabus, which may impose certain constraints on the methods used. There is also a language barrier affecting the understanding of terms like "repeated factor."

takercena
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Help me to solve this simple problem
find a if
fx = 2x^3 - ax^2 + 2x - 7
has a repeated factor
 
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Use a trial and error method. Find a value of 'x' that would make f(x)=0 for some value of 'a'. Hint: what's f(1), and what value of 'a' corresponds?
 
any easier way? Because this is in Further math syllabus so trial and error method seems too long. Or can you show me BrendanH?
 
If x0 is a double root and x1 is a different root, then your polynomial can be written 2(x- x0)2(x- x1). Multiply that out compare to the given polynomial.

"Multiple root" might mean a triple root: 2(x- x0)3. Multiply that out and compare to the given polynomial.
 
Have you tried Rational Roots Theorem and synthetic division? Picking rational roots to try will not immediately involve 'a'. The synthetic division will certainly involve 'a'. The use of synthetic division either gives remainder of zero for a root or a nonzero remainder if a choice is not a root.
 
The trouble with rational roots and factoring etc, is that the double root doesn't look like it's rational. Try this. If x is a double root then f(x)=0 and f'(x)=0. Solve those simultaneously. You wind up having to solve a cubic.
 
What is repeated factor? Does it mean that it has two or more same roots? (Its translation problem)
 
Physicsissuef said:
What is repeated factor? Does it mean that it has two or more same roots? (Its translation problem)

Yes, it has two or more of the same root.
 
symbolipoint said:
Have you tried Rational Roots Theorem and synthetic division? Picking rational roots to try will not immediately involve 'a'. The synthetic division will certainly involve 'a'. The use of synthetic division either gives remainder of zero for a root or a nonzero remainder if a choice is not a root.

Why would you think that this equation will have a rational root?
 

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