Question 523322
Recall that, if z is a root of a polynomial, then x-z is a factor of the polynomial.


Hence, we can write our new polynomial equation as (x - (ar + b))(x - (as + b)) = 0. Expanding the LHS, this yields


*[tex \LARGE x^2 - x(as + b) - x(ar + b) + (ar + b)(as + b) = 0]


*[tex \LARGE \Rightarrow x^2 - x(2b + ar + as) + (a^2rs + ab(r+s) + b^2) = 0]


Recall that by Vieta's formulas, r+s = -b/a and rs = c/a. We can substitute these expressions in:


*[tex \LARGE \Rightarrow x^2 - x(2b + a(-\frac{b}{a})) + (a^2(\frac{c}{a}) + ab(-\frac{b}{a}) + b^2) = 0]


*[tex \LARGE \Rightarrow x^2 - bx + (ac - b^2 + b^2) = 0]


b^2 cancels, so you will see that this equation is the same as in answer choice B.