MHB Mastering Factoring Polynomials: Solving Challenging Equations with Confidence

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SUMMARY

The discussion focuses on advanced techniques for factoring polynomials, specifically the expressions $3x^3-13x^2+23x-21$ and $6x^3-2x^2-13x-6$. The Rational Root Theorem was successfully applied to the first polynomial, revealing a root at $x=\frac{7}{3}$, which facilitated synthetic division. This process resulted in the factorization of the polynomial into $(3x-7)(x^2-2x+3)$. The second polynomial remains unsolved in the discussion, indicating a need for further exploration of factoring techniques.

PREREQUISITES
  • Understanding of the Rational Root Theorem
  • Proficiency in synthetic division
  • Familiarity with polynomial factorization techniques
  • Basic algebraic manipulation skills
NEXT STEPS
  • Explore advanced polynomial factoring methods, including grouping and the use of the Factor Theorem
  • Learn about synthetic division in greater depth, including its applications in polynomial long division
  • Research the application of the Rational Root Theorem to higher-degree polynomials
  • Practice solving challenging polynomial equations using various factoring techniques
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Students, educators, and mathematics enthusiasts looking to enhance their skills in polynomial factorization and problem-solving techniques in algebra.

NotaMathPerson
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After successfully solving tons of problems about factoring polynomials I've buit up quite a confidence in myself. But it did not last long when I stumbled upon these problems

Factor
$3x^3-13x^2+23x-21$
$ 6x^3-2x^2-13x-6$

None of the techniques I knew were able to help me solve these problems. Kindly assist me in this matter. Thanks much!
 
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Hello and welcome to MHB, NotaMathPerson! (Wave)

Let's walk through the first expression, and then you can try the second. Let's let:

$$f(x)=3x^3-13x^2+23x-21$$

Using the Rational Root Theorem, by trial and error, we find:

$$f\left(\frac{7}{3}\right)=0$$

So, using synthetic division, we find:

$$\begin{array}{c|rr}& 3 & -13 & 23 & -21 \\ \frac{7}{3} & & 7 & -14 & 21 \\ \hline & 3 & -6 & 9 & 0 \end{array}$$

So, we may now state:

$$f(x)=3x^3-13x^2+23x-21=\left(x-\frac{7}{3}\right)\left(3x^2-6x+9\right)=(3x-7)\left(x^2-2x+3\right)$$
 

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