How do you factor x^3-7x^2+7x+15?

I've tried grouping, but it doesn't seem to work for me.

Thanks in advance!!

Comments

  • Let f(x) = x^3-7x^2+7x+15

    By inspection (ie: guessing): f(-1) = -1 - 7 - 7 + 15 =0

    By the factor theorem is f(-1) = 0 then (x+1) is a factor of f(x).

    x^3-7x^2+7x+15 = (x+1)(Ax^2+Bx+C) where A, B and C are constants to be found either by long division or by comparing coefficients. I will be doing the latter as it's easier to show.

    Coefficients of x^3: 1 = A

    Coefficients of x^0 (constant term): 15 = C

    Coefficients of x^2: -7 = A + B , B = -8

    Thus f(x) = (x+1)(x^2-8x+15)

    We need to now see if that quadratic factors and it will if the discriminant is a complete square: b^2-4ac = 8^2 - 4*1*15 = 64-60 = 4 so this does factor:

    and so the factored expression is f(x) = (x+1)(x-3)(x-5)

  • x= -1 is a root.

    Divide by (x+1)

    x^3-7x^2+7x+15 = (x+1) (x^2 - 8x +15) = (x+1)(x-3)(x-5)

  • Yea grouping doesn't work for this. You are going to need to use synthetic division for this, so do that. Try the possibilities (p/q). I'll give you a hint. One of the factors is (x+1).

    The answer is in this link, so use it to check your answer.

    http://www.wolframalpha.com/input/?i=x^3-7x^2%2B7x...

  • x³ - 7x² + 7x + 15

    x ( x² - 7x + 7) + 15

    Maybe the binomial way:

    x ( x² - 7x + 12,25 - 12,25 + 7 ) +15

    x [ (x - 3,5)² - 5,25] + 15

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