The potential zeros have numerators which are divisors of the last term in absolute value, 21 and denominators which are divisors of the leading coefficient 1. So all potential divisors are:Try 1 1 | 1 0 -21 -20 | 1 1 -20 1 1 -20 -40 That left remainder -40, not 0, so we try the next one, -1 -1 | 1 0 -21 -20 | -1 1 20 1 -1 -20 0 That left remainder 0, so -1 is a zero, and (x+1) is a factor. The other three numbers on the bottom row give us the coefficients of the other factor, (1x2-1x-20). So f(x) factors as We can further factor the quadratic in the second parentheses: The zeros of f(x) are found by setting each factor = 0 x+1=0 ; x-5=0; x+4=0 x=-1; x=5 x=-4 The graph shows that these three zeros are at the 3 x-intercepts: Edwin