#### CHECK OUT SHMOOP'S FREE STUDY TOOLS: Essay Lab | Math Shack | Videos

# Common Core Standards: Math

#### The Standards

# High School: Algebra

### Arithmetic with Polynomials and Rational Expressions HSA-APR.B.2

**2. Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).**

Before introducing remainders of polynomials to your students, be sure they remember what a remainder is. No, not a reindeer. A *remainder*.

For example, tell them to divide 13 by 4. If they've gotten this far in math, this really shouldn't be a problem for them. Most of them will give you the answer 3.25. Instead, backtrack a few years of schooling and tell them to write their answer as "3 remainder 1." It's the same thing, only for third graders.

If we divide two integers, sometimes they make another integer (6 ÷ 2 = 3), and other times they have remainders (13 ÷ 4 = 3 remainder 1). A remainder of 0 means that the second number is a factor of the first number. For instance, 2 is a factor of 6 because we can multiply 2 by an integer to get 6.

Once they've gotten over the intense déjà-vu, slowly and gently explain to them that polynomials are the same way. If dividing polynomial *p*(*x*) by *x* – *a* has a remainder of 0, we'll know that *x* – *a* is a factor of *p*(*x*). In other words, *p*(*x*) = *q*(*x*) × (*x* – *a*) where *q*(*x*) is a polynomial or an integer.

A remainder of 0 also means that if we substitute *a* for *x*, we'll end up with *p*(*a*) = 0 regardless of what *q*(*a*) is. That's because *p*(*a*) = *q*(*a*) × (*a* – *a*) = *q*(*a*) × 0. Hopefully, by now your students know that anything multiplied by zero is zero.

If *p*(*x*) divided by *x* – *a* has a remainder, we can write the equation as *p*(*x*) = *q*(*x*) × (*x* – *a*) + *r*(*x*), where *r*(*x*) is the remainder. Sometimes *r*(*x*) is a number and other times it's another polynomial. If we set *x* = *a* in this situation, we'll end up with *p*(*a*) = *r*(*a*). Would you look at that.

Essentially, any polynomial *p*(*x*) can be written as a product of (*x* – *a*) and some quotient *q*(*x*), plus the remainder *p*(*a*).

*p*(*x*) = *q*(*x*) × (*x – a*) + *p*(*a*)

When *p*(*a*) = 0, (*x – a*) is a factor of *p*(*x*).

Students should know how to perform all these calculations and rearrangements. For their answers to be at least somewhat reasonable, they should be comfortable with factoring polynomials, finding remainders, and of course, division. You'd think that last part goes without saying, but you never know.

If students find these tasks particularly confusing, we suggest practicing with integers. That way, students can perform these same calculations within their comfort zones. Plus, they can check their answers, too. For instance,

^{8}⁄_{6} = 1 remainder 2

Converting this into the formula, we get 8 = 1(6) + 2 because 8 is like *p*(*x*), 1 is like *q*(*x*), and 2 is like *p*(*a*). Make sense? Now get them to try some polynomials instead.