Hi Mike,

Can you give me some help with question 10 in the Polynomials section of PWN (pg 149)? I plug in x=-3, per your suggestion, and I get 41 as the answer on the left-hand side of the equation and 3(-3) – 13 + a on the right. That means 41 = -22 + a, or a = 63. What am I missing? Thanks!

Sure! Here’s the question:

The key to this is recognizing that the original equation you’re getting is in quotient-remainder form: 3x-13 is the quotient with a remainder of a. Once you recognize that, you have a few options. You can actually do the division to find the remainder, you can multiply the whole equation by x+3 and solve for a, or you can use the remainder shortcut (since you’re only asked for the remainder in this question) by plugging x=-3 into only the original polynomial.

The shortcut: When a polynomial p(x) is divided by x-k, where k is a constant, you can find the remainder simply by finding p(k). In this case, that means we can find the remainder when 3x^2-4x+2 is divided by x+3 simply by simplifying 3(-3)^2-4(-3)+2, which as you note comes out to 41.

Note that we actually can’t plug in x=-3 in the original equation: we’d get division by zero on both sides.

I’m not going to do the whole long division here, but because I know that shortcut is confusing, I want to stress that just solving the equation for a is also a decent way to go. To begin, multiply everything by x+3.

    \begin{align*}\dfrac{3x^2-4x+2}{x+3}&=3x-13+\dfrac{a}{x+3}\\3x^2-4x+2&=(3x-13)(x+3)+a\\3x^2-4x+2&=3x^2+9x-13x-39+a\\3x^2-4x+2&=3x^2-4x-39+a\\2&=-39+a\\41&=a\end{align*}

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