Answer
(a) $(x-1)(x-2)(x+2)(x-5)$
(b) $f(x)=\frac{(x+4)(x+1)(x-3)(x-5)}{(x-1)(x-2)(x+2)(x-5)}$
Work Step by Step
(a) Step 1. Use the known zeros $1,2$ with synthetic division as shown in the figure.
Step 2. Use the quotient to find the rest zeros: $x^2-3x-10=0$ or $(x+2)(x-5)=0$ which gives $x=-2,5$
Step 3. Thus the numerator can be factored as $(x-1)(x-2)(x+2)(x-5)$
(b) We have $f(x)=\frac{(x+4)(x+1)(x-3)(x-5)}{(x-1)(x-2)(x+2)(x-5)}$
