There are several ways to find the zeros of the function x^4–18x^2+12x+80. You can graph the equation and see where it crosses the x axis.The graph crosses at x=-4,-2. This means that -4, and -2 are roots, and (x+4) and (x+2) are factors. I like synthetic division the best when it comes to factoring. I teach synthetic division when we get to problems like this, so I will assume that you are familiar with it.First take the coefficients of the equation x^4–18x^2+12x+80 I have to usex^4 +0 x^3 –18x^2+12x+80 : 1 0 -18 12 80. I will use -4 first.If you are not supposed to graph it, then you can use synthetic division. It can take longer because you have to “guess “ at the factors using an educated guess from the factors of =/- p/q . Where p is the last coefficient and q is the first coefficient.In this case we need to find the factors of 80/1 . These are 1,2,4,5,8,10,16,20,40,80. So possible roots are +/- (1,2,4,5,8,10,16,20,40,80). Students usually start with the smaller numbers and work up, So I would try 1 first then 2 etc.Now another way to find the factors is to use 10 as x and to see what the answer is, then try to factor the results. f(x)=x^4–18x^2+12x+80 f(10)=10,000–1800+120+80=8400I can factor 8400 into 100 *6 *14, or (10+0) *(10+0)*(10–4)*(10+4). 0 is not a factor from above, but +/- 4 is. So I could try those two numbers. Looking at the possible roots using 10 as my x, I would need to look at 9,11, 8,12, 0,20 -4,26, etc