Factor the polynomial f(x) = x^4 - 3x^3 - 7x^2 + 27x - 18 given 3 and -3 are zeros

Answer:     (x+3)(x-1)(x-2)(x-3)scroll the attachment images if you're on mobile as there are  three images that show my entire answer with the proper formatting.======    f(x) = x⁴ - 3x³ - 7x² + 27x - 18if we're given that 3 and -3 are zeros, then, by the factor theorem, these two binomials must be factors of this polynomial:     x - (3), and    x - (-3), or x + 3we can first divide f(x) by x + 3. set up synthetic division. the top row has all the coefficients of the dividend. the value, a, of the binomial divisor in the form x-a is placed on the top left. here, the divisor is x + 3 = x - (-3), so we put -3 in the top left.    in the first column, bring down the 1. then multiply that 1 with the number in the top left to get -3. after that, move that product of -3 under the next column.    add that column up. since (-3) + (-3) = -6, we put -6 under that column.repeat this process until the end.    according to synthetic division, x⁴ - 3x³ - 7x² + 27x - 18  divided by x+3 has, reading the right to left, a quotient polynomial with    remainder: 0    constant: -6    coefficient on x: 11    coefficient on x²: -6    coefficient on x³: 1therefore,      which rearranges to    we have factored out one factor. since we are given that 3 is a zero, we can divide  x³  - 6x² + 11x - 6 by x-3 using synthetic division    according to synthetic division, x³  - 6x² + 11x - 6  divided by x-3 has, reading the right to left, a quotient polynomial with    remainder: 0    constant: 2    coefficient on x: -3    coefficient on x²: 1therefore, x³  - 6x² + 11x - 6 turns into (x² - 3x + 2)(x-3)  thus    x² - 3x + 2 can be factored using quadratic methods. two numbers that multiply to get +2 (constant) and add  to get -3 (coeff. on x) are -2 and -1. sox²  -  3x  +  2 factors into (x-2) and (x-1). therefore      the complete factored form of f is      (x+3)(x-1)(x-2)(x-3)