SOLUTION: Can you please find two examples of prime polynomials? Given a polynomial such as x^2 + 4x + 8 how would I decide if this is factorable or not? What logic would you use to co

Question 280572: Can you please find two examples of prime polynomials?
Given a polynomial such as x^2 + 4x + 8 how would I decide if this is factorable or not? What logic would you use to come up with your answer? What about a more complicated polynomial such as 3x^2 + 12x + 4? How would you decide if this is prime or not?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
Neither of your equations are factorable even though this solver might say they are. This solver considers all real numbers factorable, whereas we normally only consider the equations factorable if there are whole numbers in the factors.
use the quadratic formula

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -16 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -16 is + or - .

The solution is

Here's your graph:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=96 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.367006838144548, -3.63299316185545. Here's your graph:

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?

To find these two numbers, we need to list all of the factors of (the previous product).

Factors of :

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .

1*12 = 12
2*6 = 12
3*4 = 12
(-1)*(-12) = 12
(-2)*(-6) = 12
(-3)*(-4) = 12

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	12	1+12=13
2	6	2+6=8
3	4	3+4=7
-1	-12	-1+(-12)=-13
-2	-6	-2+(-6)=-8
-3	-4	-3+(-4)=-7

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

===============================================================

Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

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