how can you tell when a quadratic such as 3n^2+9n+6 has multiple factors?

Answer 1

if you take any quadratic in the form ax^2 + bx + c

if b^2 > 4ac then it has multiple factors
in this case 81 > 72
so it has multiple factors

the factors in this case are (3x + 6)(x + 1)

Answer 2

Andrew, ALL non-trivial quadratics have multiple factors, 2 in fact. You will learn methods to find these efficiently, recognizing some special forms, and then how to factor equations that are fairly easy to factor, and then how to factor anything that is a linear quadratic in one variable.

In this case, 3 can be factored out of each term, since to factor the quadratic, it is set to zero.
Then we have n^2+3n+2. A well-practiced student will recognize that the numbers to provide 1 n^2 are always 1 and 1, and that for a positive "n" term, the other numbers will be 1 and 2. Then we write (n+1)(n+2).

Answer 3

You can figure this out easily by the discriminant of the quadratic formula. The discriminant is:

b²-4ac

Let me show you that the form of the equation that can be used for this formula is ax²+bx+c. Your equation is in this form, so we can use it. We see that a=3, since 3 is in front of the x²(n² in this case), b=9, because 9 is in front of the x (n in this case) and finally c=6, as you can see.
Now, if the discriminant is any real number other than zero, the quadratic has multiply factors. Only when the discriminant equals zero does the quadratic have only one factor.
Now, let's plug in the numbers:

9²-4(3*6)

Further simplification produces:

9²-4(18)

Even further simplification produces:

18-4(18)

Even further simplification produces:

18-72=-54.

Therefore, since it's negative, it has two imaginary roots. If it had been positive, it would have two real roots.

There you go. I'm glad I could help. :)

Answer 4

Check the discriminant of the quadratic formula.

If b^2 - 4ac is a positive number, then the quadratic has either 1 repeating root or two roots.

Otherwise, both roots are complex.

Answer 5

all coeffs divisible by 3 so
3(n^2 + 3n+2)
can immediately see
3(n+1)(n+2)

if you cou'dnt immediately see can fall back on the quadratic formula.