Dividing Polynomials
Dividing polynomials works like long division with numbers. Dividing \(P(x)\) by a divisor gives a quotient and a remainder, written \(P(x)=A(x)Q(x)+R(x)\) for a divisor \(A(x)\). In NSW Year 10 Advanced maths you divide by linear expressions such as \(x+2\) and \(2x-1\), and also by quadratics (NESA outcome MA5‑POL‑P‑01).
You will use long division to find quotients and remainders, write \(P(x)\) as divisor times quotient plus remainder, and use a zero remainder to factorise a cubic fully. You will find a length from an area and a width, divide by quadratics, and work backwards from a quotient and remainder to the polynomial or its coefficients. It is part of Year 10 Advanced Logarithms and Polynomials.
Every question with a fully worked solution.
- Dividing Polynomials - Video - Polynomials: division Watch
Theory
Polynomial long division works like dividing whole numbers: divide the leading terms, multiply back, subtract, bring down, repeat. This Year 10 Advanced maths page covers the quotient and remainder, missing powers, quadratic and non-monic divisors, and the statement \(P(x) = D(x)Q(x) + R(x)\).
When \(157\) is divided by \(6\), \(157\) is the dividend, \(6\) the divisor, \(26\) the quotient and \(1\) the remainder: \(157 = 6 \times 26 + 1\).
Polynomials divide the same way. Dividing \(P(x)\) by \(D(x)\) gives a quotient \(Q(x)\) and a remainder \(R(x)\) with \(P(x) = D(x)\,Q(x) + R(x)\).
The division stops when the remainder has a lower degree than the divisor. A linear divisor leaves a constant; a quadratic divisor leaves a linear expression or a constant.
If the remainder is \(0\), the divisor is a factor of \(P(x)\), and the quotient is the other factor.
This comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
| Divisor | Quotient degree (for a quartic \(P\)) | Remainder looks like |
|---|---|---|
| \(x - a\) or \(2x + 1\) | \(3\) | a number |
| \(x^2 + bx + c\) | \(2\) | \(px + q\) |
How to do polynomial long division
- Set up: write both polynomials in descending powers. Fill every missing power with a \(0\) term.
- Divide the leading term of what is left by the leading term of the divisor. Write the answer above the column with the same power.
- Multiply the whole divisor by that term and write it underneath, lined up by powers.
- Subtract (change every sign of the line you are subtracting), then bring down the next term.
- Repeat until what is left has a lower degree than the divisor. That is the remainder.
- Check: expand \(D(x)Q(x) + R(x)\), or substitute a value of \(x\).
Other question types
- Hence factorise: if the remainder is \(0\), factorise the quotient further (Example 4).
- Find \(P(x)\): multiply the divisor by the quotient and add the remainder.
- Find unknown coefficients: build \(D(x)Q(x) + R(x)\), expand, and equate coefficients with the given \(P(x)\).
Divide the leading term, multiply back, subtract, bring down the next term, and repeat.
| \(2x^2\) | \(+x\) | \(+6\) | ||
| \(x - 2\) | \(2x^3\) | \(-3x^2\) | \(+4x\) | \(-7\) |
| \(-\) | \(2x^3\) | \(-4x^2\) | ||
| \(x^2\) | \(+4x\) | |||
| \(-\) | \(x^2\) | \(-2x\) | ||
| \(6x\) | \(-7\) | |||
| \(-\) | \(6x\) | \(-12\) | ||
| \(5\) |
\(2x^3 \div x = 2x^2\), then \(x^2 \div x = x\), then \(6x \div x = 6\). The last line, \(5\), has a lower degree than \(x - 2\), so stop.
| \(Q(x)\) | \(=\) | \(\boldsymbol{2x^2 + x + 6}\) |
| \(R\) | \(=\) | \(\boldsymbol{5}\) |
| \(P(x)\) | \(=\) | \((x - 2)(2x^2 + x + 6) + 5\) |
Check with \(x = 2\): \(P(2) = 16 - 12 + 8 - 7 = 5\), the remainder.
There is no \(x^3\) term and no \(x\) term. Write them in as \(0x^3\) and \(0x\) so every column lines up.
| \(x^3\) | \(-x^2\) | \(-4x\) | \(+4\) | ||
| \(x + 1\) | \(x^4\) | \(+0x^3\) | \(-5x^2\) | \(+0x\) | \(+3\) |
| \(-\) | \(x^4\) | \(+x^3\) | |||
| \(-x^3\) | \(-5x^2\) | ||||
| \(-\) | \(-x^3\) | \(-x^2\) | |||
| \(-4x^2\) | \(+0x\) | ||||
| \(-\) | \(-4x^2\) | \(-4x\) | |||
| \(4x\) | \(+3\) | ||||
| \(-\) | \(4x\) | \(+4\) | |||
| \(-1\) |
| \(Q(x)\) | \(=\) | \(\boldsymbol{x^3 - x^2 - 4x + 4}\) |
| \(R\) | \(=\) | \(\boldsymbol{-1}\) |
Watch the signs when subtracting: \(0x^3 - x^3 = -x^3\), and \(0x - (-4x) = 4x\). Check: at \(x = -1\), \(1 - 5 + 3 = -1\), the remainder.
Write in \(0x^2\). Each step now uses three columns of the divisor.
| \(x^2\) | \(+3x\) | \(+1\) | |||
| \(x^2 - x + 2\) | \(x^4\) | \(+2x^3\) | \(+0x^2\) | \(-x\) | \(+6\) |
| \(-\) | \(x^4\) | \(-x^3\) | \(+2x^2\) | ||
| \(3x^3\) | \(-2x^2\) | \(-x\) | |||
| \(-\) | \(3x^3\) | \(-3x^2\) | \(+6x\) | ||
| \(x^2\) | \(-7x\) | \(+6\) | |||
| \(-\) | \(x^2\) | \(-x\) | \(+2\) | ||
| \(-6x\) | \(+4\) |
| \(Q(x)\) | \(=\) | \(\boldsymbol{x^2 + 3x + 1}\) |
| \(R(x)\) | \(=\) | \(\boldsymbol{-6x + 4}\) |
Stop when the remainder has a lower degree than \(x^2 - x + 2\): a linear remainder is allowed.
Divide each leading term by \(2x\), not \(x\): \(2x^3 \div 2x = x^2\).
| \(x^2\) | \(+x\) | \(-6\) | ||
| \(2x - 1\) | \(2x^3\) | \(+x^2\) | \(-13x\) | \(+6\) |
| \(-\) | \(2x^3\) | \(-x^2\) | ||
| \(2x^2\) | \(-13x\) | |||
| \(-\) | \(2x^2\) | \(-x\) | ||
| \(-12x\) | \(+6\) | |||
| \(-\) | \(-12x\) | \(+6\) | ||
| \(0\) |
The remainder is \(0\), so \((2x - 1)\) is a factor and the quotient is the other factor.
| \(2x^3 + x^2 - 13x + 6\) | \(=\) | \((2x - 1)(x^2 + x - 6)\) |
| \(\) | \(=\) | \(\boldsymbol{(2x - 1)(x + 3)(x - 2)}\) |
Common pitfalls
Frequently asked questions
How do you divide polynomials using long division?
Write both in descending powers with zeros for missing terms. Divide the leading terms, write the result on top, multiply the divisor by it, subtract, bring down the next term, and repeat until the remainder has a lower degree than the divisor.
What is the remainder when dividing polynomials?
What is left at the end of the division, with a degree lower than the divisor. Dividing by \(x - 3\) always leaves a number; dividing by a quadratic leaves at most a linear expression.
How do you write P(x) = D(x)Q(x) + R(x)?
After dividing, write the dividend equal to divisor \(\times\) quotient \(+\) remainder. For example \(2x^3 - 3x^2 + 4x - 7 = (x - 2)(2x^2 + x + 6) + 5\).
Why do you put 0x in polynomial division?
A zero term holds the place of a missing power, so that like powers stay in the same column when you multiply and subtract.
What does a remainder of zero mean?
The divisor is a factor of the polynomial. The polynomial equals the divisor times the quotient, so you can factorise it further using the quotient.