Math Problem Statement
Find root using Rational root theorem x⁴-5x³+6x²+4x-8
Solution
The Rational Root Theorem helps to find the possible rational roots of a polynomial equation. For the polynomial , the Rational Root Theorem states that any rational root, when expressed as a fraction , is a factor of the constant term , and is a factor of the leading coefficient .
Step 1: List Factors
Factors of the constant term (-8):
Factors of the leading coefficient (1):
Step 2: Possible Rational Roots
Using the Rational Root Theorem, the possible rational roots are:
Step 3: Test the Possible Rational Roots
We will use synthetic division or direct substitution to test these possible roots.
Testing : So, is not a root.
Testing : So, is a root.
Step 4: Factor the Polynomial
Since is a root, we can factor out .
Perform synthetic division of by :
-1 & 1 & -5 & 6 & 4 & -8 \\ & & -1 & 6 & -12 & 8 \\ \hline & 1 & -6 & 12 & -8 & 0 \\ \end{array}$$ The quotient is: $$x^3 - 6x^2 + 12x - 8$$ ### Step 5: Repeat the Process Now we factor $$x^3 - 6x^2 + 12x - 8$$. Again, list the factors of the constant term (-8) and the leading coefficient (1). Possible roots are again: $$\pm 1, \pm 2, \pm 4, \pm 8$$ **Testing $$x = 1$$:** $$P(1) = 1^3 - 6 \cdot 1^2 + 12 \cdot 1 - 8 = 1 - 6 + 12 - 8 = -1 \neq 0$$ So, $$x = 1$$ is not a root. **Testing $$x = -1$$:** $$P(-1) = (-1)^3 - 6 \cdot (-1)^2 + 12 \cdot (-1) - 8 = -1 - 6 - 12 - 8 = -27 \neq 0$$ So, $$x = -1$$ is not a root. **Testing $$x = 2$$:** $$P(2) = 2^3 - 6 \cdot 2^2 + 12 \cdot 2 - 8 = 8 - 24 + 24 - 8 = 0$$ So, $$x = 2$$ is a root. Perform synthetic division of $$x^3 - 6x^2 + 12x - 8$$ by $$x - 2$$: $$\begin{array}{r|rrrr} 2 & 1 & -6 & 12 & -8 \\ & & 2 & -8 & 8 \\ \hline & 1 & -4 & 4 & 0 \\ \end{array}$$ The quotient is: $$x^2 - 4x + 4$$ ### Step 6: Solve the Remaining Quadratic Equation Solve $$x^2 - 4x + 4 = 0$$: $$(x - 2)^2 = 0 \implies x = 2$$ ### Step 7: Collect All Roots The polynomial $$P(x) = x^4 - 5x^3 + 6x^2 + 4x - 8$$ has the following roots: $$x = -1, x = 2 \text{ (with multiplicity 3)}$$ ### Summary The roots of the polynomial are: $$x = -1, \quad x = 2, \quad x = 2, \quad x = 2$$ Would you like more details or have any questions? Here are some related questions: 1. How is the Rational Root Theorem derived? 2. What are the applications of the Rational Root Theorem? 3. Can the Rational Root Theorem be used for polynomials with irrational or complex coefficients? 4. What is synthetic division and how is it used in polynomial division? 5. How do you determine the multiplicity of a root? 6. What are other methods to find the roots of polynomials besides the Rational Root Theorem? 7. How does the Rational Root Theorem relate to the Fundamental Theorem of Algebra? 8. How can graphing a polynomial help in finding its roots? **Tip:** When using synthetic division, always write down each step clearly to avoid mistakes in calculations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Polynomial equations
Rational Root Theorem
Synthetic division
Formulas
-
Theorems
Rational Root Theorem
Suitable Grade Level
Grades 10-12
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