An SAT Math micro-topic under Polynomial and other nonlinear graphs (Advanced Math). Free to read — no account needed.
Dividing usually leaves a remainder. When you divide 17 by 5, you get 3 with 2 left over — that 2 is the remainder. In the same way, dividing a polynomial p(x) by (x−c) leaves a remainder.
The Remainder Theorem gives a shortcut: the remainder is simply p(c). So instead of doing long division, you just substitute c into the polynomial. For p(x)=x2+1 divided by (x−3), the remainder is p(3)=9+1=10.
The number c is whatever makes (x−c) equal to zero. For (x−2), c=2. For (x+1), rewrite it as (x−(−1)), so c=−1, and the remainder is p(−1).
Sometimes you are told the remainder and must find a missing number. If dividing p(x)=x2+kx+1 by (x−1) leaves a remainder of 5, then p(1)=5. So 1+k+1=5, which gives k=3.
Worked examples
What is the remainder when p(x)=x2+2x is divided by (x−4)? By the shortcut, the remainder is p(4). So p(4)=42+2(4)=16+8=24.
What is the remainder when p(x)=x2−5 is divided by (x+1)? Here (x+1)=(x−(−1)), so c=−1 and the remainder is p(−1). So p(−1)=(−1)2−5=1−5=−4.
Dividing p(x)=x2+3x+k by (x−2) leaves a remainder of 1. Find k. The remainder is p(2)=4+6+k=10+k. Setting 10+k=1 gives k=−9.
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