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What is the completely factored form of f(x) = 6x3 - 13x2 - 4x + 15?
Solution:
The given polynomial is
f(x) = 6x3 - 13x2 - 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.
6x3 - 13x2 - 4x + 15 / x + 1
= 6x2 + ( -19x2 - 4x + 15 / x + 1)
= 6x2 + ( -(19x - 15) (x + 1) / (x + 1)
= 6x2 - 19x + 15
Dividing the polynomial by (x + 1) the quotient we get is 6x2 - 19x + 15
6x3 - 13x2 - 4x + 15 = (x + 1)(6x2 - 19x + 15)
= (x + 1)(6x2 - 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) = 6x3 - 13x2 - 4x + 15?
Summary:
The completely factored form of f(x) = 6x3 - 13x2 - 4x + 15 is (x + 1)(2x - 3)(3x - 5).
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