One root is x = 2. I got that by knowing that any rational root must be + or 1, 2, 3, 4, 6 or 12 (the integer factors of 12, the last term), and by trying it out. Therfore (x-2) is one of the factors of the polynomial. Dividing that into the original fourth order polynomial will tell you that the other factor is
x^3 - 5x^2 +8x -6. See if you can factor that by trying + or - 1,2,3,4 are zeros of that function. Negative numbers don't work becasue all of the erms are negative. Try x = 3. It works. So x-3 is another factor. Divide that into x^3 - 5x^2 +8x -6 and you get
x^2 -2x +2 for the remaining factor. Since b^2 - 4ac is negative, that tells you that the remaining roots are complex. Calculate them with the quadratic equation.
The final factored form is
(x-2)(x-3)(x^2 -2x +2)
Given f(x)=x^4-7x^3+18x^2-22x+12
A: list all possible rational zeros
B: Graph f(x)
C: Find all zeros
D: show f(x) in factured form
thanks I can't figure it out. Can skip B... thanks
1 answer