how do you solve these problems: 2n(squared) - 11n + 15 and
-3n(squared) - 75

Question

anonymous · Answer

are you dealing with imaginary numbers?

anonymous · Answer

no i am dealing with factoring polynomials

anonymous · Answer

so 2n^2-11n+15 = 2n^2 -6n-5n+15

anonymous · Answer

thank you so much!!

anonymous · Answer

-3n^2-75=0--------> n^2=-25

anonymous · Answer

thank you soo much!!

anonymous · Answer

you are welcome. note that for the second equation, you will have imaginary roots.

radar · Answer

Since we are tasked to factor the polynomials, take a close look at the second equation.$$-3n ^{2}-75$$

radar · Answer

It is a factor problem, extract the -3 which is present in both terms.$$-3(n ^{2}+25)$$

radar · Answer

Thats is about as far as you can go with this.

anonymous · Answer

That's true, unless the right hand side is specified in the equation.

radar · Answer

I didn't notice any but if it is zero, we would have the situation that you mentioned and probably doehrman hasn't got into complex number yet.

anonymous · Answer

yes, you are right