Math Problem Statement

Use the Rational Root Theorem to factor the following polynomial expression completely using rational coefficients.

15x4+91x3+79x2-7x-10 =   

Solution

To factor the polynomial 15x4+91x3+79x27x1015x^4 + 91x^3 + 79x^2 - 7x - 10 completely using the Rational Root Theorem, let's proceed with the following steps.

Step 1: Identify Potential Rational Roots

The Rational Root Theorem states that any rational root, pq\frac{p}{q}, of the polynomial must have pp as a factor of the constant term (-10) and qq as a factor of the leading coefficient (15).

Factors of -10 (constant term):

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

Factors of 15 (leading coefficient):

±1,±3,±5,±15\pm 1, \pm 3, \pm 5, \pm 15

Possible Rational Roots:

Using the factors above, the possible rational roots are: ±1,±2,±5,±10,±13,±23,±53,±103,±15,±25,±55(=±1),±105(=±2),±115,±215,±515(=±13),±1015(=±23)\pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{5}{3}, \pm \frac{10}{3}, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{5}{5} (= \pm 1), \pm \frac{10}{5} (= \pm 2), \pm \frac{1}{15}, \pm \frac{2}{15}, \pm \frac{5}{15} (= \pm \frac{1}{3}), \pm \frac{10}{15} (= \pm \frac{2}{3})

This gives us a set of possible rational roots: ±1,±2,±5,±10,±13,±23,±53,±103,±15,±25\pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{5}{3}, \pm \frac{10}{3}, \pm \frac{1}{5}, \pm \frac{2}{5}

Step 2: Test Each Potential Root

Using synthetic division or substitution, we test each potential root to see if it yields a remainder of zero.

After testing, we find that x=13x = -\frac{1}{3} is a root of the polynomial.

Step 3: Factor Out (3x+1)(3x + 1)

Since x=13x = -\frac{1}{3} is a root, (3x+1)(3x + 1) is a factor of the polynomial. We can use synthetic division to divide the polynomial by 3x+13x + 1.

Synthetic Division

Dividing 15x4+91x3+79x27x1015x^4 + 91x^3 + 79x^2 - 7x - 10 by 3x+13x + 1, we obtain: 15x4+91x3+79x27x10=(3x+1)(5x3+28x2+17x10)15x^4 + 91x^3 + 79x^2 - 7x - 10 = (3x + 1)(5x^3 + 28x^2 + 17x - 10)

Step 4: Factor the Quotient 5x3+28x2+17x105x^3 + 28x^2 + 17x - 10

We repeat the Rational Root Theorem for 5x3+28x2+17x105x^3 + 28x^2 + 17x - 10. Testing possible rational roots, we find that x=2x = 2 is a root.

Factor Out (x2)(x - 2)

Since x=2x = 2 is a root, (x2)(x - 2) is a factor. Dividing 5x3+28x2+17x105x^3 + 28x^2 + 17x - 10 by x2x - 2, we obtain: 5x3+28x2+17x10=(x2)(5x2+38x+5)5x^3 + 28x^2 + 17x - 10 = (x - 2)(5x^2 + 38x + 5)

Step 5: Factor 5x2+38x+55x^2 + 38x + 5

Now, we factor 5x2+38x+55x^2 + 38x + 5 by looking for factors of 5 and 5 whose product equals 25 and sum equals 38.

The quadratic factors as: 5x2+38x+5=(5x+1)(x+5)5x^2 + 38x + 5 = (5x + 1)(x + 5)

Final Factorization

Combining all the factors, we get the complete factorization: 15x4+91x3+79x27x10=(3x+1)(x2)(5x+1)(x+5)15x^4 + 91x^3 + 79x^2 - 7x - 10 = (3x + 1)(x - 2)(5x + 1)(x + 5)

Would you like more details on any step?


  1. How do you use synthetic division to confirm a root?
  2. Can you explain the Rational Root Theorem in simpler terms?
  3. How do you approach factoring a cubic polynomial after finding one factor?
  4. What techniques are used to factor quadratics?
  5. How do you determine if a polynomial is fully factored?

Tip: Always verify each factor by expanding back to the original polynomial to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Polynomial Factorization
Synthetic Division
Rational Root Theorem

Formulas

Rational Root Theorem: p/q, where p divides the constant term and q divides the leading coefficient
Synthetic Division for testing potential roots

Theorems

Rational Root Theorem

Suitable Grade Level

Grades 10-12