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# What is the completely factored form of f(x) = 6x^{3} - 13x^{2} - 4x + 15?

**Solution:**

The given polynomial is

f(x) = 6x^{3} - 13x^{2} - 4x + 15

Assume x = -1

f(-1) = 6(-1)^{3} - 13(-1)^{2} - 4(-1) + 15

f(-1) = -6 -13 + 4 + 15

f(-1) = 0

Using the factor theorem, (x + 1) is a factor of the given polynomial.

6x^{3} - 13x^{2} - 4x + 15 / x + 1

= 6x^{2} + ( -19x^{2} - 4x + 15 / x + 1)

= 6x^{2} + ( -(19x - 15) (x + 1) / (x + 1)

= 6x^{2} - 19x + 15

Dividing the polynomial by (x + 1) the quotient we get is 6x^{2} - 19x + 15

6x^{3} - 13x^{2} - 4x + 15 = (x + 1)(6x^{2} - 19x + 15)

= (x + 1)(6x^{2} - 10x - 9x + 15)

= (x + 1)[2x(3x - 5) - 3(3x - 5)]

So we get,

= (x + 1)(2x - 3)(3x - 5)

Therefore, the completely factored form is (x + 1)(2x - 3)(3x - 5).

## What is the completely factored form of f(x) = 6x^{3} - 13x^{2} - 4x + 15?

**Summary:**

The completely factored form of f(x) = 6x^{3} - 13x^{2} - 4x + 15 is (x + 1)(2x - 3)(3x - 5).

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