01 · Choose the active variable
If the problem says to divide in x, then x is the active variable and y behaves like a fixed number stored in a box.
x² + (y+2)x + 2y has x-coefficients 1, y+2, 2y
1Divide leading terms.
2Multiply the divisor.
3Subtract the whole row.
4Bring down.
02 · A complete example
Divide (6x²+(3y+4)x+2y) by (3x+2).
First, (6x²÷3x=2x). Multiplying gives (6x²+4x). Subtracting leaves (3yx+2y). Next, (3yx÷3x=y). Multiplying gives (3yx+2y), so the remainder is zero.
(6x²+(3y+4)x+2y) ÷ (3x+2) = 2x+y
03 · Remainders may contain y
Dividing (x²+yx+1) by (x+1) gives quotient (x+y−1) and remainder (2−y). This is finished because the remainder contains no x.
P(x,y)=(ax+b)Q(x,y)+R(y)
The zero of ax+b is x=−b/a, so the fast remainder rule is:
R(y)=P(−b/a,y)
04 · Common traps
| Trap | Better habit |
|---|---|
| Mixing x and y | Say “divide in x; y is a coefficient.” |
| Skipping a missing power | Write a placeholder such as 0x². |
| Dividing by only x | Use the full leading term ax. |
| Changing one sign | Subtract the entire product in parentheses. |
05 · Your turn
Divide x²+8x+15 by x+3.
x+5.
Find c so x−4 divides x²+cx−20.
Set P(4)=0: 16+4c−20=0, so c=1.
Find the remainder of 2x³+yx+7 divided by 2x+1.
Substitute x=−½: R(y)=27/4−y/2.