Factor completely. if the polynomial is prime, state this.
-6x^2-21x-9
I am confused about how to factoring.
Thanks!
-6x^2-21x-9
first, take out the -3
-3(2x^2+7x+3)
-3(x +3)(2x+1)
Somebody had the same question earlier today.
http://www.jiskha.com/display.cgi?id=1459354736
Thanks for the hints.
To factor a polynomial completely, we want to express it as the product of irreducible factors.
To begin factoring the given polynomial -6x^2 - 21x - 9, we can start by looking for common factors among the coefficients. In this case, all the coefficients are divisible by -3, so we can factor out -3:
-3(2x^2 + 7x + 3)
Now we have a trinomial, 2x^2 + 7x + 3, to factor further.
To factor a trinomial of the form ax^2 + bx + c, we need to find two binomials that multiply to give the trinomial. In this case, we are looking for binomials of the form (px + q)(rx + s).
We want to find p, q, r, and s such that:
(p * r) = 2 (the coefficient of x^2)
(p * s) + (q * r) = 7 (the coefficient of x)
(q * s) = 3 (the constant term)
From these equations, we can see that the factors of 2 are 1 and 2. The factors of 3 are 1 and 3. We need to find the combination of these factors that satisfies the second equation.
Trying out different combinations, we find that p = 1, q = 3, r = 1, and s = 2 work:
(2x + 3)(x + 1)
Therefore, the completely factored form of the given polynomial is:
-3(2x + 3)(x + 1)
Note: The factored form is not prime because it can be further reduced.