Originally Posted by
HallsofIvy ??? What was "part a"? The quadratic formula?
Perhaps:
b) The fact that r and s satisfy $\displaystyle ax^2+ bx+ c= 0$ means that (x- r) and (x- s) are factors. Since just multiplying them together gives [tex]x^2- (r+s)x+ rs, to get the "a" multiplying $\displaystyle x^2$, we must multiply that product by a: a(x-r)(x-s)= ax^2- a(r+s)x+ ars= ax^2+ bx+ c.
c) Use the quadratic formula (or completing the square) to find the roots of each equation. Then use the fact, from (b) above, that the polynomial can be factored as a(x-r)(x-s) where a is the leading coefficient (1 in the first example, 5 in the second) and r and s are the solutions to the equations with the polynomial set to 0.