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

f(x) = 6x³ - 13x² - 4x + 15
f(x) = 6x³ - 19x² + 6x² + 15x - 19x + 15
f(x) = 6x² - 19x² + 15x + 6x² - 19x + 15
f(x) = x(6x²) - x(19x) + x(15) + 1(6x²) - 1(19x) + 1(15)
f(x) = x(6x² - 19x + 15) + 1(6x² - 19x + 15)
f(x) = (x + 1)(6x² - 19x + 15)
f(x) = (x + 1)(6x² - 10x - 9x + 15)
f(x) = (x + 1)(2x(3x) - 2x(5) - 3(3x) + 3(5))
f(x) = (x + 1)(2x(3x - 5) - 3(3x - 5))
f(x) = (x + 1)(2x - 3)(3x - 5)

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

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erato erato · Answer 2 · 2019-04-23T19:15:11+02:00

Answer:

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

Step-by-step explanation:

f(x) = 6x³ - 13x² - 4x + 15

This could also be written as:

f(x) = 6x³ - 19x² + 6x² + 15x - 19x + 15

f(x) = 6x² - 19x² + 15x + 6x² - 19x + 15

f(x) = x(6x² - 19x + 15) + 1(6x² - 19x + 15)

f(x) = (x + 1)(6x² - 19x + 15)

f(x) = (x + 1)(6x² - 10x - 9x + 15)

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

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

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