Question

find the real zeros of the given functions, use the real zeros to factor f. f(x)=-2x4power+9x3power+2x2power-39x+18

Answer 1

I found f(3) = 0 , x-3 is a factor
then found f(-2) = 0 , so x+2 is a factor
divided both of those and got
-(2x^2-7x+3)
which factored further to -(x-3)(2x-1)

so f(x) = -(x-3)^2 (x+2)(2x-1)

with zeros at x= -2, 1/2 and 3 , 3 being a double root

Answer 2

234357

Answer 3

y = -2 x^4 + 9 x^3 + 2 x^2 - 39 x + 18

make a table of x and y
x _____ y

-2 _____ 0 oh wow

-1 _____ 48

0 ______ 18

+1 ______ -12 so a zero between 0 and 1

well, at least we had a zero at x = -2 so (x+2) is a factor
so divide this mess by (x+2)
so I have
(x+2)(-2x^3+13x^2-24x+9)
well, try .5 or divide by (2x-1)
because we have that 2 there and we know we have a zero around .5
It works and I have
(x+2)(2x-1)(-x^2+6x-9)
well 9 does not factor by much so try 3
-(3)^2 + 6(3) -9 = 18 -18 = 0
so I have
(x+2)(2x-1)(x-3)(x-3) but we want negative those (zero has either sign) so multiply by -1
(-1)(x+2)(2x-1)(x-3)(x-3)