factor p(x)=6x^(5)-11x^(4)-15x^(3)+45x^(2)-31x+6
6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6 = 0
rational roots will be ±1,±2,±3, ...
So try the easy ones first. You can easily see that x=1 will work, since the coefficients add to zero.
6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6 = (x-1)(6x^4-5x^3-20x^2+26x-6)
Again, x=1 will work, and we have
6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6 = (x-1)^2 (6x^3+x^2-19x+6)
Now a little synthetic division yields
(x-1)^2 (x+2)(6x^2-11x+3)
and that factors to
(x-1)^2 (x+2)(2x-3)(3x-1)
oobleck is a star player
To factor the polynomial p(x) = 6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6, you can use various factoring techniques such as factoring by grouping or using the rational root theorem to find the roots.
Let's start by checking if there are any rational roots using the rational root theorem. According to the rational root theorem, the possible rational roots of the polynomial are the factors of the constant term divided by the factors of the leading coefficient.
In this case, the constant term is 6, and the leading coefficient is 6. So, the possible rational roots are ± factors of 6 divided by ± factors of 6.
The factors of 6 are ±1, ±2, ±3, and ±6. Dividing these factors by ±1, ±2, ±3, and ±6 gives us a list of potential rational roots: ±1, ±2, ±3, ±6.
Next, to check which of these potential roots are actual roots of the polynomial, you can use synthetic division or plug them into the polynomial and see if they give a result of zero.
Let's try one potential root to demonstrate the process. Let's use x = 1 as an example.
Performing synthetic division with the divisor x - 1:
1 | 6 -11 -15 45 -31 6
| 6 -5 -20 25 -6
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6 -5 -20 25 -6 0
Since the remainder is zero when dividing by x - 1, it means that (x - 1) is a factor of p(x). Therefore, we have factored out (x - 1) from p(x) as:
p(x) = (x - 1)(6x^4 - 5x^3 - 20x^2 + 25x - 6)
Now, we can repeat the process with the new polynomial 6x^4 - 5x^3 - 20x^2 + 25x - 6 to find more factors.
By testing the potential rational roots again, either by synthetic division or plugging them into the polynomial, we find that another root is x = 2.
Performing synthetic division with the divisor x - 2:
2 | 6 -5 -20 25 -6
| 12 14 -12 26
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6 7 -6 13 20
The remainder is not zero, so (x - 2) is not a factor.
Continue this process of checking potential roots until all factors have been found, or until a manageable equation is left that is easier to factor further.
Note: In some cases, polynomial factors may involve complex numbers or other mathematical techniques. However, for this specific polynomial, following the steps outlined above should help you in factoring it.