Finding the Completely Factored Form of the Polynomial
To factor the polynomial 6x³ + 13x² + 4x – 15, we can follow a systematic approach that involves several steps:
Step 1: Use the Rational Root Theorem
The Rational Root Theorem suggests that any rational solution (root) of the polynomial equation can be expressed as a fraction p/q, where p is a factor of the constant term (-15) and q is a factor of the leading coefficient (6).
The factors of -15 are: ±1, ±3, ±5, ±15
The factors of 6 are: ±1, ±2, ±3, ±6
Listing all possible rational roots, we have: ±1, ±1/2, ±1/3, ±1/6, ±3, ±5, ±15, ±3/2, ±5/2, ±15/2
Step 2: Test Possible Roots
We can test these values to see which, if any, are roots of the polynomial. By substituting and evaluating:
f(1) = 6(1)³ + 13(1)² + 4(1) - 15 = 8 (not a root)
f(-1) = 6(-1)³ + 13(-1)² + 4(-1) - 15 = -12 (not a root)
f(3) = 6(3)³ + 13(3)² + 4(3) - 15 = 0 (is a root!)
Step 3: Synthetic Division
Since x = 3 is a root, we can use synthetic division to divide the polynomial by (x - 3):
3 | 6 13 4 -15
| 18 93 300
------------------
6 31 4 0
The result of the synthetic division is 6x² + 31x + 5. So, we have:
6x³ + 13x² + 4x – 15 = (x – 3)(6x² + 31x + 5)
Step 4: Factor the Quadratic
Now we need to factor the quadratic polynomial 6x² + 31x + 5. We look for two numbers that multiply to 6 * 5 = 30 and add to 31. These numbers are 30 and 1:
6x² + 30x + 1x + 5
= 6x(x + 5) + 1(x + 5)
= (6x + 1)(x + 5)
Final Factored Form
Substituting this back in, we get:
6x³ + 13x² + 4x – 15 = (x – 3)(6x + 1)(x + 5)
Conclusion
The completely factored form of the polynomial 6x³ + 13x² + 4x - 15 is:
(x - 3)(6x + 1)(x + 5)