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

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    ProtonsTalk
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      What is the completely factored form of f(x) = 6x3 – 13x2 – 4x + 15?

      • This topic was modified 2 years ago by ProtonsTalk.
      • This topic was modified 2 years ago by ProtonsTalk.
      • This topic was modified 1 year, 7 months ago by ProtonsTalk.
      #20895
      ProtonsTalk
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        The completely factored form of f(x) = 6x3 – 13x2 – 4x + 15 is (x + 1)(2x – 3)(3x – 5).

        We know that after quadratic equations, we don’t have any direct method to find roots or factors. So for cubic and above equations we have to find roots by trial and error until we reach to quadratic equations.

        So, Let us start with an assumption, say x = -1

        Substituting x = -1 in the equation, we get

        => 6(-1)3 – 13(-1)2 – 4(-1) + 15

        => -6 -13 + 4 + 15 = 0

        So, f(-1) = 0 which means -1 is a root and x + 1 is a factor of the given polynomial.

        Dividing the polynomial by x+1, we get

        => 6x3 – 13x2 – 4x + 15 / (x + 1)

        Splitting -13x2 as 6x2 – 19x2

        => 6x3 + 6x2 +  -19x2 – 4x + 15 / (x + 1)

        Cancelling x+1

        => 6x2(x+1) / (x+1) + (-19x2 – 4x + 15 / (x + 1))

        => 6x2 + (-19x2 – 4x + 15 / (x + 1))

        Factorizing the other part and cancelling x+1

        => 6x2 + ( – (19x  – 15) (x + 1)  / (x + 1)

        => 6x2 – 19x + 15

        So, the factorized form till now is as follows

        6x3 – 13x2 – 4x + 15 = (x + 1)(6x2 – 19x + 15)

        Now, we can directly factorize the quadratic factor

        => 6x2 – 19x + 15

        => 6x2 – 10x – 9x + 15

        => 2x(3x – 5) – 3(3x – 5)

        => (2x – 3)(3x – 5)

        So the complete factored form of 6x3 – 13x2 – 4x + 15 is

        => (x + 1)(6x2 – 19x + 15)

        => (x + 1)(2x – 3)(3x – 5)

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