what is the completely factored form of f(x) = 6x3 – 13x2 – 4x + 15?
(x + 1)(6x2 – 19x + 15)
(x + 1)2(2x – 3)
(x + 1)(3x – 2)(5x – 3)
(x + 1)(2x – 3)(3x – 5)

Respuesta :

We have to determine the complete factored form of the given polynomial

[tex]f(x)=6x^3-13x^2-4x+15[/tex].

Let x= -1 in the given polynomial.

So, [tex]f(-1)= 6(-1)^3-13(-1)^2-4(-1)+15 = -6-13+4+15 = 0[/tex]

So, by factor theorem

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

So, dividing the given polynomial by (x+1), we get quotient as [tex]6x^2-19x+15[/tex].

So, [tex]6x^3-13x^2-4x+15[/tex] = (x+1)[tex]6x^2-19x+15[/tex].

= [tex](x+1)(6x^2-10x-9x+15)[/tex]

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

= [tex](x+1)(2x-3)(3x-5)[/tex] is the completely factored form of the given polynomial.

Option D is the correct answer.

Answer:

D) (x + 1)(2x – 3)(3x – 5)

Step-by-step explanation: