First see if any root that is an integer factor of 6 works in solving
r^3 + 6r^2 + 11r +6 = 0
Try these prime factors of 6:
+ or - 1,2,3, or 6.
+ roots won't work because all polynomial terms would be positive.
-1 works; therefore r+1 is a factor.
Divide (r+1) into r^3+6r^2+11r+6 for the other factor, using polynomial long division. You should get r^2 + 5r + 6.
That can also be factored. Do that and you're done.
I don't know how to factor this polynomial.
r^3+6r^2+11r+6
Please help.
7 answers
That is confusing. We aren't learning it that way.
There is not way to make factoring cubic polynomials easy. The answer to your problem is (x-1)(x-2)(x-3). I did it the quickest way.
Hmmmm. If you are not learning it that way, you are missing a great lesson.
If you are not learning it in a way similar to the method drwls showed you, I would be very curious to know how "they" are teaching you to factor cubic polynomials
I made a mistake in my second post. I mixed up the polynomial equation roots (-1, -2, and -3) with the factors. The factors are (r+1)(r+2)(r+3). The polynomial is zero whenever a factor is zero. Sorry about the confusion.
Solve the following quadratic equation by factoring x^2-6x-16=0