State the possible number of imaginary zeros of g(x)= x^4+3x^3+7x^2-6x-13 this is how far Ive gotten:there are 4 zeros,3 positive and 1 negative If you've found four real zeroes, then there are no imaginary zeroes. You can also
question The square root of a negative number is imaginary. That is you cannot find, for example, sqrt(-49) . Why? my answer Yes you can, however this is aginst the normal operations non the less we are dealing with iminiganary
1)Use synthetic subsitution to find f(-3)for f(x)=x^4-4x^3+2x^2-4x+6 answer= -15 2)One factor of x^3+2x^2-11x-12 is x+4. find the remaining factors answer= x-1;x-3 3)Which describes the number and type of roots of the equation
find the intervals on which f(x) is increasing and decreasing along with the local extrema. f(x)=x^4 + 18x^2 I took the derivative and got: f'(x)= 4x^3 + 36x When I set this to zero, I get the imaginary number 3i. I can't get test
"Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the two real roots and the four imaginary roots of this equation." I used synthetic division to get the real roots 2 and -1, but I can't figure out how to get the
i^6 = (i^2)^3 = (-1)^3 = -1 The square, not the square root of -1 is 1. :) B.t.w., can you prove that -1 times -1 is 1? Hint, try to prove first that for any number X: -1 times X equals -X How do you find the square root of -1?
Which describes the number and type of roots of the equation x^2-625=0? A)1 real root, 1 imaginary root B)2 real roots, 2 imaginary roots C)2 real roots D)4 real roots I went with A sqrt x^2= x (which would be the imaginary) and
The function 5x^2+2x-1's domain is all real numbers, according to the answer key. However, I don't know how to factor this to prove it. Could you show me? The function the sq. root of y-10 has a domain of all numbers so that y is