SAT Mathematics Practice Test 7

In general, a binomial of the form x + f, where f is a constant, is a factor of a polynomial when the remainder of dividing the polynomial by x + f is 0. Let R be the remainder resulting from the division of the polynomial $\mathit{P}\left(\mathit{x}\right)=\mathit{a}{\mathit{x}}^3+\mathit{b}{\mathit{x}}^2+\mathit{c}\mathit{x}+\mathit{d}$ by x + 1. So the polynomial P(x) can be rewritten as P(x) = (x + 1)q(x) + R, where q(x) is a polynomial of second degree and R is a constant. Since –1 is a root of the equation P(x) = 0, it follows that P(–1) = 0. Since P(–1) = 0 and P(–1) = R, it follows that R = 0. This means that x + 1 is a factor of P(x).