Question

What are the possible number of positive, negative, and complex zeros of f(x) = –2x3 – 5x2 + 6x + 4

Answer 1

Since the coefficients are given, there is only one possible number of complex zeros. Find the roots of the equation.

Do you know calculus?

Answer 2

no i dont know calc

Answer 3

Complex roots always come in pairs. So if (a+bi) is a root, so is (a-bi). This is because when you rewrite the polynomial to include (x-(a+bi))*(x-(a-bi)), it is expands to become x^2 - 2ax + (a^2-b^2), and those pesky i's go away!

As it turns out, this is the ONLY way the i's go away, and because there are NO i's in your original equation, then the imaginary roots MUST occur in pairs.

So knowing this, what possible number of complex roots could there be?

Answer 4

yea i need people to show me

Answer 5

it will go this way
-6-5(X)^2+6X+4=0
-5(X)^2+6X-2=0

Since the value of D is -4 that is -ve we will have two complex roots.
x=(-6+sqrt(D))/(2*(-5)) & x=(-6-sqrt(D))/(2*(-5)) that is
x=3/5-i/5 & x=3/5+i/5

you see since the graph doesn't cuts x axis it has no solution