Question

State how many imaginary and real zeros the function has. f(x) = x3 + 5x2 - 28x - 32

Answer 1

Two numbers that are easy to check as possible zeros for a polynomial are 1 and -1.
f(1) = sum of coefficients of f = 1+5-28-32=-54, so 1 isn't a zero.
f(-1)= sum of coefficients when odd powers of x switch signs = -1+5+28-32=0, so -1 is a zero.

Now you divide f(x) by the factor (x+1) (since -1 is a zero, (x+1) is a factor). You can do this with long division or synthetic division. Either way, you get this quadratic: x^2+4x-32. This factors into (x-4)(x+8).

So, f(x)=(x+1)(x-4)(x+8), which means f(x) has 3 real zeros.

Answer 2

thank you very much ur a lifesaver ive been stuck on this for hours ! :o