6(t^2 + 12)

how can i tell if this is factored completly

6(t^2 + 12)

how can i tell if this is factored completly

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Hi Christian:

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Your question gets at the heart of what factoring is. Every factoring problem you ever get will require you to look at the problem and decide if there is any way to factor it. The only easy way to tell is if you know how to factor well enough to see that it cannot be factored any further.

In this case I’m sure you are wondering if t^2 + 12 can be factored even more. With enough practice you will see that the general form t^2 + c, where c is just a constant greater than 0 (like 12), can never be factored with integer coefficients. That’s because any numbers you choose that multiply together to make 12, will also make a t term. For instance (t +2)(t+6) makes an 8t. You get t^2+8t+12. Any time you have t^2 plus a positive number, that is going to happen.

Negative numbers are a different story. Sometimes they can be factored. t^2 – 16 can be factored to (t+4)(t-4). That works because the +4t cancels out the -4t.

I mentioned this was the only easy way. You can tell this problem is factored all the way (over the set of integers) by plugging it into something called the quadratic equation. I won’t go into that here, but it shows that you cannot factor this problem with integers, you can’t even factor it with the whole set of real numbers! You have use imaginary numbers!

1. There is formula "ax^{2} + bx + c = a(x-x_{1})(x-x_{2}) , x_{1
}and x_{2 }are roots of equation ax2 + bx + c = 0 "

2. You can not factorize it in the set of real numbers.

3. In set of complex numbers the solution is:

6(x^{2} + 12) = 0 -----> x^{2} + 12 = 0 -----> x^{2} = -12 ------> x = ± √(-12) ------> x = ± 2* i* √3

The final answer is: 6(x + 2

Jed L.

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