F(x)=x^3+4x^2+x-6 what are the real zeros

Happy New Year from MrBillDoesMath!

   1, -2, -3

Discussion:

The real roots of the polynomials are factors of the last term (6) divided by the first term (1). The factors of 6/1 or 6 are 1,2,3, and 6. I found by substitution that x= 1 was a solution, divided the polynomial by x-1, came up with a quadratic (x^2 + 5x + 6), and found the remaining roots via the quadratic formula.

Thank you,

MrB

you answer is a

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.