Tagged: polynomial, polynomials
 This topic has 1 reply, 1 voice, and was last updated 1 year, 6 months ago by ProtonsTalk.

AuthorPosts

May 15, 2022 at 8:07 pm #20892
What is the completely factored form of f(x) = 6x^{3} – 13x^{2} – 4x + 15?
 This topic was modified 1 year, 6 months ago by ProtonsTalk.
 This topic was modified 1 year, 6 months ago by ProtonsTalk.
 This topic was modified 1 year, 2 months ago by ProtonsTalk.
May 15, 2022 at 8:46 pm #20895::The completely factored form of f(x) = 6x^{3} – 13x^{2} – 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
=> 6x^{3} – 13x^{2} – 4x + 15 / (x + 1)
Splitting 13x^{2} as 6x^{2} – 19x^{2}
=> 6x^{3} + 6x^{2} + 19x^{2} – 4x + 15 / (x + 1)
Cancelling x+1
=> 6x^{2}(x+1) / (x+1) + (19x^{2} – 4x + 15 / (x + 1))
=> 6x^{2 }+ (19x^{2} – 4x + 15 / (x + 1))
Factorizing the other part and cancelling x+1
=> 6x^{2} + ( – (19x – 15) (x + 1) / (x + 1)
=> 6x^{2} – 19x + 15
So, the factorized form till now is as follows
6x^{3} – 13x^{2} – 4x + 15 = (x + 1)(6x^{2} – 19x + 15)
Now, we can directly factorize the quadratic factor
=> 6x^{2} – 19x + 15
=> 6x^{2} – 10x – 9x + 15
=> 2x(3x – 5) – 3(3x – 5)
=> (2x – 3)(3x – 5)
So the complete factored form of 6x^{3} – 13x^{2} – 4x + 15 is
=> (x + 1)(6x^{2} – 19x + 15)
=> (x + 1)(2x – 3)(3x – 5)

AuthorPosts
 You must be logged in to reply to this topic.