Polynomials and Factoring
Polynomial Long Division
The division algorithm for polynomials, laid out beside the arithmetic long division it copies, with placeholders, remainders and the quotient-plus-remainder form.
What this page covers
- Set out a polynomial long division with correct placeholders
- Carry the algorithm through to a remainder of lower degree than the divisor
- Write the result in quotient-plus-remainder form and check it
- Recognise a zero remainder as a statement that the divisor is a factor
The arithmetic it copies
Polynomial long division is the school method for dividing numbers, transplanted. The source notes make the parallel explicit by doing an arithmetic division first, and the comparison is worth keeping in view throughout.
- Dividend
- The number or polynomial being divided — here 183
- Divisor
- What it is divided by — here 17
- Quotient
- How many times it goes — here 10
- Remainder
- What is left over — here 13
The defining property is that the remainder is smaller than the divisor. For numbers 'smaller' means smaller in value; for polynomials it means lower in degree. That is the only substantive change in the transplant.
Arithmetic warm-up — 68437 div 314
| Step | Working | Reasoning |
|---|---|---|
| 1 | 314 into 684 goes 2 | Put 2 in the quotient |
| 2 | 2 × 314 = 628; subtract | Leaves 56, bring down 3 |
| 3 | 314 into 563 goes 1 | Put 1 in the quotient |
| 4 | 1 × 314 = 314; subtract | Leaves 249, bring down 7 |
| 5 | 314 into 2497 goes 7 | Put 7 in the quotient |
| 6 | 7 × 314 = 2198; subtract | Leaves 299 |
The algorithm for polynomials
- Write both polynomials in standard form, descending by degree, with a zero placeholder for every missing power.
- Divide the leading term of the dividend by the leading term of the divisor. That quotient term goes on top.
- Multiply the whole divisor by that quotient term and write the product underneath, aligned by degree.
- Subtract. The leading terms must cancel; if they do not, the quotient term was wrong.
- Bring down the next term and repeat from step 2.
- Stop when the degree of the remainder is strictly less than the degree of the divisor.
Step 4 is where signs are lost. You are subtracting the whole product, so every term of it changes sign. Writing the product and then explicitly negating it on the next line is slower and much safer than trying to subtract in the head.
Worked example: a linear divisor
Divide x3 - 5x2 + 7x - 12 by x - 4
| Quotient term | Multiply divisor by it | Subtract, leaving |
|---|---|---|
| x2 | x3 - 4x2 | -x2 + 7x |
| -x | -x2 + 4x | 3x - 12 |
| +3 | 3x - 12 | 0 |
Each quotient term comes from dividing the current leading term by x: first x3 div x = x2, then -x2 div x = -x, then 3x div x = 3.
(x - 4)(x2 - x + 3) = x3 - x2 + 3x - 4x2 + 4x - 12 = x3 - 5x2 + 7x - 12. Correct.
A remainder of zero is a statement worth reading carefully: it says x - 4 is a factor of the dividend. That observation is the whole content of the factor theorem.
Worked example: a quadratic divisor and a non-zero remainder
Divide 2x4 - x2 - 2 by x2 + 2x - 1
The dividend is missing its x3 and x terms, so write it in full before starting:
| Quotient term | Multiply divisor by it | Subtract, leaving |
|---|---|---|
| 2x2 | 2x4 + 4x3 - 2x2 | -4x3 + x2 + 0x |
| -4x | -4x3 - 8x2 + 4x | 9x2 - 4x - 2 |
| +9 | 9x2 + 18x - 9 | -22x + 7 |
The remainder -22x + 7 has degree 1, which is below the divisor's degree 2, so the division stops.
(x2 + 2x - 1)(2x2 - 4x + 9) = 2x4 - x2 + 22x - 9. Adding the remainder -22x + 7 gives 2x4 - x2 - 2. Correct.
Notice how much work the placeholders did. Without the 0x3 the second subtraction would have aligned -4x3 against -x2, and everything after that would have been wrong while still looking orderly.
The two standard forms of the answer
The second form is the more useful of the two. It contains no fractions, it is the form the check is done in, and it is the form from which the remainder theorem is derived in one line.
| Form | Example |
|---|---|
| Fractional | x3 - 5x2 + 7x - 12x - 4 = x2 - x + 3 |
| Product | x3 - 5x2 + 7x - 12 = (x - 4)(x2 - x + 3) |
| General, with remainder | f(x) = q(x)g(x) + r(x), ° r < ° g |
Using division to factor a cubic
Division is the second half of the standard method for factoring a cubic: find one root by trial, then divide it out to leave a quadratic, which any of the usual methods can handle.
Worked example — f(x) = 2x3 + 5x2 - 14x + 3
The typed notes test x = 32:
Dividing 2x3 + 5x2 - 14x + 3 by 2x - 3:
| Quotient term | Multiply divisor by it | Subtract, leaving |
|---|---|---|
| x2 | 2x3 - 3x2 | 8x2 - 14x |
| +4x | 8x2 - 12x | -2x + 3 |
| -1 | -2x + 3 | 0 |
The quadratic does not factor over the integers, so finish with the formula: x = -4 ± √202 = -2 ± √5.
The trial step is not guesswork if it is done properly. Which values are worth testing, and why, is set out in The Remainder and Factor Theorems.
Common mistakes
| Mistake | Consequence | Prevention |
|---|---|---|
| Omitting placeholder terms | Columns misalign; every later step is wrong | Write the dividend in full, with 0 coefficients |
| Adding instead of subtracting | The leading term fails to cancel | Negate the product on its own line before combining |
| Stopping while the remainder still has high degree | An incomplete quotient | Compare degrees explicitly before stopping |
| Dividing the whole leading coefficient into the wrong term | Wrong quotient term | Divide leading term by leading term only |
| Not checking | Errors survive to the final answer | q · g + r must reproduce f exactly |
| Writing the divisor out of standard order | The algorithm cannot start | Sort both polynomials by descending degree first |
Frequently asked questions
Why do I need placeholder zeros?
The algorithm aligns terms by degree. If x3 - 2 is written without its missing x2 and x terms, the subtraction at each stage lines up the wrong columns and every subsequent step is wrong. Writing x3 + 0x2 + 0x - 2 costs nothing and removes the risk entirely.
When do I stop dividing?
When the degree of what remains is strictly less than the degree of the divisor. Dividing by a quadratic, you stop once the remainder is linear or constant. Dividing by a linear factor, you stop at a constant.
Does the remainder have to be zero?
No. A zero remainder is the special case that says the divisor is a factor. A non-zero remainder is a perfectly good answer and is written as the fractional part of the quotient-plus-remainder form.
Is there a faster method?
For a linear divisor x - a, yes: the remainder alone can be got from f(a) by the remainder theorem, with no division at all. Long division is still needed when the quotient itself is wanted, or when the divisor has degree two or more.
Source. Handwritten teaching notes, Week 6, pages 6-7, with the arithmetic parallel and the cubic factorisation from the typed supplementary notes on long division.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.