Math Problem Statement

The function f, of, x, equals, x, cubed, plus, 2, x, squared, minus, 10, x, minus, 20f(x)=x3+2x2−10x−20 has at least one integer root. Use the integer root theorem to find that root, then proceed to find all complex roots. (Note: roots may be integer, rational, irrational, and/or complex.)

Solution

To solve the given problem, we will follow these steps:

Step 1: Apply the Integer Root Theorem

The integer root theorem states that any possible rational root of a polynomial must be a factor of the constant term divided by a factor of the leading coefficient.

The given polynomial is:

f(x)=x3+2x210x20f(x) = x^3 + 2x^2 - 10x - 20

  • The constant term is 20-20.
  • The leading coefficient (of x3x^3) is 11.

Thus, the possible integer roots are the factors of 20-20. The factors of 20-20 are:

±1,±2,±4,±5,±10,±20\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20

Step 2: Test Possible Integer Roots

Now, we'll substitute these possible integer roots into f(x)f(x) and see which one gives f(x)=0f(x) = 0.

  • For x=1x = 1:

f(1)=(1)3+2(1)210(1)20=1+21020=27(not a root)f(1) = (1)^3 + 2(1)^2 - 10(1) - 20 = 1 + 2 - 10 - 20 = -27 \quad (\text{not a root})

  • For x=1x = -1:

f(1)=(1)3+2(1)210(1)20=1+2+1020=9(not a root)f(-1) = (-1)^3 + 2(-1)^2 - 10(-1) - 20 = -1 + 2 + 10 - 20 = -9 \quad (\text{not a root})

  • For x=2x = 2:

f(2)=(2)3+2(2)210(2)20=8+82020=24(not a root)f(2) = (2)^3 + 2(2)^2 - 10(2) - 20 = 8 + 8 - 20 - 20 = -24 \quad (\text{not a root})

  • For x=2x = -2:

f(2)=(2)3+2(2)210(2)20=8+8+2020=0(a root)f(-2) = (-2)^3 + 2(-2)^2 - 10(-2) - 20 = -8 + 8 + 20 - 20 = 0 \quad (\text{a root})

So, x=2x = -2 is an integer root.

Step 3: Perform Polynomial Division

Now that we have x=2x = -2 as a root, we can factor f(x)f(x) as (x+2)(quotient)(x + 2)(\text{quotient}). To find the quotient, we perform synthetic division of f(x)f(x) by (x+2)(x + 2).

Synthetic Division:

We divide x3+2x210x20x^3 + 2x^2 - 10x - 20 by x+2x + 2.

-2 |  1   2   -10   -20
   |      -2    0     20
   ---------------------
     1   0   -10     0

The quotient is x210x^2 - 10, and the remainder is 0. Therefore, the polynomial can be factored as:

f(x)=(x+2)(x210)f(x) = (x + 2)(x^2 - 10)

Step 4: Solve for the Remaining Roots

Now, we solve the quadratic x210=0x^2 - 10 = 0 for the remaining roots.

x2=10x=±10x^2 = 10 \quad \Rightarrow \quad x = \pm \sqrt{10}

Thus, the roots are x=10x = \sqrt{10} and x=10x = -\sqrt{10}.

Final Answer:

The roots of the polynomial f(x)=x3+2x210x20f(x) = x^3 + 2x^2 - 10x - 20 are:

  • x=2x = -2 (integer root),
  • x=10x = \sqrt{10} (irrational root),
  • x=10x = -\sqrt{10} (irrational root).

Would you like more details on any of the steps or further explanation?

Related Questions:

  1. How do you perform synthetic division for higher-degree polynomials?
  2. Can the integer root theorem be applied to any polynomial?
  3. What are the conditions for a polynomial to have complex roots?
  4. How do we check for rational roots in a polynomial using the Rational Root Theorem?
  5. What is the geometric significance of irrational roots on the graph of a polynomial?

Tip:

When solving for roots of polynomials, always check for integer roots first, as they simplify the process of factorization and help you find more complex roots later.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomials
Roots of Polynomials

Formulas

f(x) = x^3 + 2x^2 - 10x - 20
Synthetic division
Quadratic equation for solving x^2 - 10 = 0

Theorems

Integer Root Theorem
Quadratic Root Theorem

Suitable Grade Level

Grades 10-12