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 P(x)=x45x3+6x2+4x8P(x) = x^4 - 5x3 + 6x^2 + 4x - 8, the Rational Root Theorem states that any rational root, when expressed as a fraction pq\frac{p}{q}, pp is a factor of the constant term 8-8, and qq is a factor of the leading coefficient 11.

Step 1: List Factors

Factors of the constant term (-8): ±1,±2,±4,±8\pm 1, \pm 2, \pm 4, \pm 8

Factors of the leading coefficient (1): ±1\pm 1

Step 2: Possible Rational Roots

Using the Rational Root Theorem, the possible rational roots are: ±1,±2,±4,±8\pm 1, \pm 2, \pm 4, \pm 8

Step 3: Test the Possible Rational Roots

We will use synthetic division or direct substitution to test these possible roots.

Testing x=1x = 1: P(1)=14513+612+418=15+6+48=20P(1) = 1^4 - 5 \cdot 1^3 + 6 \cdot 1^2 + 4 \cdot 1 - 8 = 1 - 5 + 6 + 4 - 8 = -2 \neq 0 So, x=1x = 1 is not a root.

Testing x=1x = -1: P(1)=(1)45(1)3+6(1)2+4(1)8=1+5+648=0P(-1) = (-1)^4 - 5 \cdot (-1)^3 + 6 \cdot (-1)^2 + 4 \cdot (-1) - 8 = 1 + 5 + 6 - 4 - 8 = 0 So, x=1x = -1 is a root.

Step 4: Factor the Polynomial

Since x=1x = -1 is a root, we can factor out (x+1)(x + 1).

Perform synthetic division of P(x)P(x) by x+1x + 1:

-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.

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Math Problem Analysis

Mathematical Concepts

Polynomial equations
Rational Root Theorem
Synthetic division

Formulas

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Theorems

Rational Root Theorem

Suitable Grade Level

Grades 10-12