The Remainder Theorem is not something that you will use many times when taking the SAT, but it has shown up on a couple of problems in the practice tests that have been released in *The Official SAT Study Guide*.

The Remainder Theorem says that if you divide some polynomial *p*(*x*) by the linear factor *x* – *a*, the remainder that you get is equal to *p*(*a*). For example, consider the problem below where the polynomial *p*(*x*) = *x**² – *2*x* – 11 is divided by *x* – 5 using either Long Division or Synthetic Division.

The remainder in both cases is 4. The Remainder Theorem says that if the remainder when you divide by *x* – 5 is 4, then it is also true that *p*(5) = 4. It is easy to check that this is indeed true.

*p*(*x*) = *x**² – *2*x* – 11

*p*(5) = 5² – 2(5) – 11

*p*(5) = 25 – 10 – 11

*p*(5) = 4

There is also a relationship between the polynomial *p*(*x*), the divisor (*x *– *a*), the quotient *q*(*x*) and the remainder *r*. Specifically

*p*(*x*) = *q*(*x*)⋅(*x* – *a*) + r

Using the example above, you can write

*p*(*x*) = (*x* + 3)(*x* – 5) + 4

*p*(*x*) = *x**² – *5*x* + 3*x* – 15 + 4

*p*(*x*) = *x**² – *2*x* – 11

OK, let’s try a few problems. When you’re done, scroll down for the answers.

1. A student used division to divide a polynomial *p*(*x*) by *x* – 3 and got a remainder of 5. Which of the following statements about *p*(*x*) must be true?

A) *x* – 3 is a factor of *p*(*x*).

B) *x* – 5 is a factor of *p*(x)

C) The value of *p*(-3) is 5.

D) The value of *p*(3) is 5.

2. When a polynomial *p*(*x*) is divided by* x* – 4, the quotient is another polynomial *q*(*x*), and the remainder is -6. Which of the following must be true of *p*(*x*)?

A) *p*(*x*) = (*x* – 4)(*x* – 6)

B) *p*(x) = (*x* – 4)(*x* + 6)

C) *p*(*x*) = *q*(*x*)(*x* – 4) – 6

D) *p*(*x*) = *q*(*x*)·(*x* – 4) + 6

3. When a polynomial *p*(*x*) is divided by *x* + 3, the quotient *q*(*x*) is *x* – 8 and the remainder *r* is 5. Which of the following is the polynomial *p*(*x*)?

A) *p*(*x*) = *x*^{2} – 5*x* – 29

B) *p*(*x*) = *x*^{2} – 5*x* – 19

C) *p*(*x*) = *x*^{2} + 5*x* + 19

D) *p*(*x*) = *x*^{2} + 5*x* + 29

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(Scroll down for answers)

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1. D

2. C

3. B

*p*(*x*) = (*x* + 3)(*x* – 8) + 5

*p*(*x*) = *x*² – 8*x* + 3*x* – 24 + 5

*p*(*x*) = *x**² –* 5*x* – 19

If you have questions about these problems or anything else related to the SAT, send us an email at info@cardinalec. com.