Why substitution leaves the remainder
Dividing a polynomial by x − a leaves a constant remainder r. In P(x) = (x − a)Q(x) + r, setting x = a makes the product term zero, leaving P(a) = r.
This avoids carrying out the full division when only the remainder is needed. For a divisor x + 2, use a = −2, not 2. The sign comes from matching x − a.
Use the factor theorem as a special case
If P(a) = 0, then x − a is a factor. If P(a) is nonzero, it is the remainder, and the linear expression is not a factor.
The basic calculator can evaluate the substituted arithmetic. It does not manipulate P(x) symbolically; replace every x with the chosen number, using parentheses around negative substitutions.
A linear divisor need not be monic
For a divisor such as 2x − 3, substitute its zero x = 3/2 to find the constant remainder. The identity P(x) = (2x − 3)Q(x) + r still leaves only r at that value. Standard synthetic division tables usually assume a divisor x − a, so account for the leading coefficient if you also need the quotient.
LET’S WORK IT THROUGH
Find the remainder when x³ − 4x + 5 is divided by x − 2.
- Use a = 2.
- P(2) = 2³ − 4 × 2 + 5.
- 8 − 8 + 5 = 5.
THE ANSWERThe remainder is 5.
YOUR TURN
A small check for understanding.
Is x + 1 a factor of x² + 3x + 2?
Reveal the answer
Yes. P(−1) = 1 − 3 + 2 = 0.
Now try your own numbers.
The basic calculator includes input guidance and the limits of its supported calculations. Use it alongside the method above.
Open Basic calculator ↗A note on precision: preserve exact fractions where possible and round numerical results at the end. Found an unclear step? Prepare a correction report.