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

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

234357

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)

To find the real zeros of the given function f(x) = -2x^4 + 9x^3 + 2x^2 - 39x + 18, we can use a method called the Rational Root Theorem and synthetic division.

1. Start by listing all possible rational roots, which are in the form of p/q, where p is a factor of the constant term (18 in this case), and q is a factor of the leading coefficient (-2 in this case). In this example, the factors of 18 are ±1, ±2, ±3, ±6, ±9, and ±18. The factors of -2 are ±1 and ±2. So, the possible rational roots are ±1, ±2, ±3, ±6, ±9, and ±18, divided by ±1 and ±2.

2. Use synthetic division to test each possible root until you find one that gives a remainder of zero. Start by selecting one of the possible roots and divide it by the polynomial using synthetic division. For example, let's start with the root x = 1:

1 | -2 9 2 -39 18
| -2 7 9 -30
|----------------------------
-2 7 9 -30 -12

The result is -2x^3 + 7x^2 + 9x - 30 with a remainder of -12.
Since the remainder is not zero, 1 is not a root of the polynomial.

3. Repeat step 2 with the remaining possible roots until you find a remainder of zero. After testing all the possible rational roots, we find that x = 3 is a real root of the polynomial.

3 | -2 9 2 -39 18
| -6 9 33
|----------------------------
-2 3 11 -6 51

The result is -2x^2 + 3x + 11 with a remainder of 51.
Since the remainder is not zero, 3 is not a root of the polynomial either.

4. The real zeros of the polynomial f(x) = -2x^4 + 9x^3 + 2x^2 - 39x + 18 are x = 1 and x = 3.

To factor f(x), we can use the real zeros we found: x = 1 and x = 3.

f(x) = -2x^4 + 9x^3 + 2x^2 - 39x + 18

Since x = 1 is a real zero, (x - 1) is a factor.
Similarly, since x = 3 is a real zero, (x - 3) is a factor.

Using synthetic division, divide the polynomial by (x - 1) first:

1 | -2 9 2 -39 18
| -2 7 9 -30
|----------------------------
-2 7 9 -30 -12

The result is -2x^3 + 7x^2 + 9x - 30.

Now, divide this result by (x - 3):

3 | -2 7 9 -30 -12
| -6 3 36
|----------------------------
-2 1 12 6 24

The final result is -2x^2 + x + 12 with a remainder of 24.

Therefore, factored form of the given function is: f(x) = (x - 1)(x - 3)(-2x^2 + x + 12).