Math Problem Statement

x^3-4x^2+x-4 roots

Solution

To find the roots of the cubic equation x34x2+x4=0x^3 - 4x^2 + x - 4 = 0, we can first try to factor the equation or use methods like the Rational Root Theorem.

Step 1: Rational Root Theorem

The Rational Root Theorem suggests that the potential rational roots could be the factors of the constant term (4-4) divided by the factors of the leading coefficient (which is 11).

So, the possible rational roots are:
±1,±2,±4\pm 1, \pm 2, \pm 4

Step 2: Test the Possible Rational Roots

We can test these values by substituting them into the equation f(x)=x34x2+x4f(x) = x^3 - 4x^2 + x - 4.

Testing x=1x = 1:

f(1)=(1)34(1)2+(1)4=14+14=6f(1) = (1)^3 - 4(1)^2 + (1) - 4 = 1 - 4 + 1 - 4 = -6 So, x=1x = 1 is not a root.

Testing x=1x = -1:

f(1)=(1)34(1)2+(1)4=1414=10f(-1) = (-1)^3 - 4(-1)^2 + (-1) - 4 = -1 - 4 - 1 - 4 = -10 So, x=1x = -1 is not a root.

Testing x=2x = 2:

f(2)=(2)34(2)2+(2)4=816+24=10f(2) = (2)^3 - 4(2)^2 + (2) - 4 = 8 - 16 + 2 - 4 = -10 So, x=2x = 2 is not a root.

Testing x=2x = -2:

f(2)=(2)34(2)2+(2)4=81624=30f(-2) = (-2)^3 - 4(-2)^2 + (-2) - 4 = -8 - 16 - 2 - 4 = -30 So, x=2x = -2 is not a root.

Testing x=4x = 4:

f(4)=(4)34(4)2+(4)4=6464+44=0f(4) = (4)^3 - 4(4)^2 + (4) - 4 = 64 - 64 + 4 - 4 = 0 So, x=4x = 4 is a root.

Step 3: Factor the Polynomial

Since x=4x = 4 is a root, we can now factor the cubic equation by dividing x34x2+x4x^3 - 4x^2 + x - 4 by (x4)(x - 4) using synthetic division or polynomial division.

Performing synthetic division of x34x2+x4x^3 - 4x^2 + x - 4 by x4x - 4:

4 & 1 & -4 & 1 & -4 \\ & & 4 & 0 & 4 \\ \hline & 1 & 0 & 1 & 0 \\ \end{array}$$ The quotient is $$x^2 + 1$$, so the factorization is: $$x^3 - 4x^2 + x - 4 = (x - 4)(x^2 + 1)$$ ### Step 4: Solve for the Roots Now, we solve for the roots of the quadratic factor $$x^2 + 1 = 0$$. $$x^2 + 1 = 0 \quad \Rightarrow \quad x^2 = -1 \quad \Rightarrow \quad x = \pm i$$ ### Final Roots: The roots of the cubic equation $$x^3 - 4x^2 + x - 4 = 0$$ are: $$x = 4, \, x = i, \, x = -i$$ Would you like further explanation or assistance with any step? Here are some related questions you might find interesting: 1. How do we apply synthetic division to polynomials? 2. Can we factor a cubic equation using methods other than synthetic division? 3. What is the significance of complex roots in polynomials? 4. How can we find roots of a polynomial with higher degrees (degree 4, 5, etc.)? 5. Can all cubic equations be factored using simple methods like this? **Tip:** When trying to factor a cubic polynomial, always check for rational roots first using the Rational Root Theorem before moving to more advanced techniques.

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Math Problem Analysis

Mathematical Concepts

Algebra
Cubic Equations
Factorization
Complex Numbers
Synthetic Division

Formulas

Rational Root Theorem
Synthetic Division
x^2 + 1 = 0

Theorems

Rational Root Theorem

Suitable Grade Level

Grades 10-12