Long Division of Polynomials

How many x + 2 fit?

Long division of numbers asks how many times the divisor fits into the number being divided, and what is left over. Long division of polynomials asks the same question. To divide $x^2+5x+7$x² + 5x + 7 by x + 2, ask: how many lots of x + 2 fit into $x^2+5x+7$x² + 5x + 7?

$x^2+5x+7$x² + 5x + 7 is the dividend and x + 2 is the divisor. The answer has two parts, as with numbers: the quotient, how many lots fit, and the remainder, what is left over. Write both polynomials with the highest power first, the way a number is written with its largest place value first.

Divide the leading terms

Compare only the leading terms, the terms with the highest power. The leading term of the dividend is $x^2$x², and the leading term of the divisor is x. $x^2/x=x$x² / x = x, so the quotient starts with x.

This is the "divide" step of Divide, Multiply, Subtract, Bring down. In the number division 528 ÷ 12, the first question is how many 12s fit into 52; here it is how many x fit into $x^2$x².

Multiply back and subtract

Multiply: the whole divisor times the new term, $x\left(x+2\right)=x^2+2x$x(x + 2) = x² + 2x. Subtract it from the first two terms of the dividend: $\left(x^2+5x\right)-\left(x^2+2x\right)=3x$(x² + 5x) − (x² + 2x) = 3x. Bring down the next term, + 7, and what is left to divide is 3x + 7.

The $x^2$x² cancels, and that is why the quotient term was chosen by dividing the leading terms: each round removes the highest power that is left.

The division drawn as a rectangle: the side down the left is the divisor, x + 2, and the side along the top is the quotient being found. The first column, x wide, has area $x^2+2x$x² + 2x. Taking it away from $x^2+5x+7$x² + 5x + 7 leaves 3x + 7 to place.

Run the loop again

Repeat the four steps on 3x + 7. Divide the leading terms: $3x/x=3$3x / x = 3, so the quotient gains + 3. Multiply back: 3(x + 2) = 3x + 6. Subtract: (3x + 7) − (3x + 6) = 1. There are no more terms to bring down.

The 1 has no x in it, so x + 2 cannot fit into it again: the loop stops. In general, stop when what is left has a lower degree, a lower highest power, than the divisor. That leftover is the remainder.

The second column, 3 wide, has area 3x + 6. The finished rectangle is $\left(x+2\right)\left(x+3\right)=x^2+5x+6$(x + 2)(x + 3) = x² + 5x + 6, which is 1 short of $x^2+5x+7$x² + 5x + 7: the remainder.

Quotient and remainder

The quotient is x + 3 and the remainder is 1. As a single statement: $x^2+5x+7=\left(x+2\right)\left(x+3\right)+1$x² + 5x + 7 = (x + 2)(x + 3) + 1. Check by multiplying back: $\left(x+2\right)\left(x+3\right)=x^2+3x+2x+6=x^2+5x+6$(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6, and adding 1 gives $x^2+5x+7$x² + 5x + 7.

Dividing both sides by x + 2 writes the division as a quotient plus a fraction: $\frac{\left(x^2+5x+7\right)}{\left(x+2\right)}=x+3+\frac{1}{\left(x+2\right)}$(x² + 5x + 7)/(x + 2) = x + 3 + 1/(x + 2). The remainder sits over the divisor, just as $25÷12=2+\frac{1}{12}$25 ÷ 12 = 2 + 1/12.

The statement also shows what the remainder is. Put x = −2: the bracket x + 2 is 0, so the first part vanishes and $x^2+5x+7$x² + 5x + 7 equals the remainder. Indeed ${\left(-2\right)}^2+5\times \left(-2\right)+7=4-10+7=1$(−2)² + 5 × (−2) + 7 = 4 − 10 + 7 = 1. When the remainder is 0, the divisor is a factor of the dividend.

The same division with numbers

Put x = 10 into the division. The dividend $x^2+5x+7$x² + 5x + 7 becomes 100 + 50 + 7 = 157, the divisor x + 2 becomes 12, and the quotient x + 3 becomes 13. So the working should say 157 ÷ 12 = 13 remainder 1, and it does: 13 × 12 = 156, and 157 − 156 = 1.

The rounds match as well. The first round took away $x\left(x+2\right)=x^2+2x$x(x + 2) = x² + 2x, which is 10 × 12 = 120, and left 3x + 7, which is 37. The second round took away 3(x + 2) = 3x + 6, which is 36, and left 1. Polynomial long division is the same loop as number long division, with the powers of x in place of the place values.

157 ÷ 12 = 13 remainder 1. With x = 10 this is $x^2+5x+7$x² + 5x + 7 divided by x + 2: the 1 in 13 stands for 10, which is x, and the 37 left after the first round is 3x + 7.

A missing power

Divide $x^3-8$x³ − 8 by x − 2. The dividend has no $x^2$x² term and no x term, so write them in with a coefficient of 0: $x^3+0x^2+0x-8$x³ + 0x² + 0x − 8. Each power then has its own place, as the 0 in 105 holds the tens place.

Round 1: $x^3/x=x^2$x³ / x = x². Multiply back: $x^2\left(x-2\right)=x^3-2x^2$x²(x − 2) = x³ − 2x². Subtract: $\left(x^3+0x^2\right)-\left(x^3-2x^2\right)=2x^2$(x³ + 0x²) − (x³ − 2x²) = 2x², and bring down 0x to get $2x^2+0x$2x² + 0x.

Round 2: $2x^2/x=2x$2x² / x = 2x. Multiply back: $2x\left(x-2\right)=2x^2-4x$2x(x − 2) = 2x² − 4x. Subtract: $\left(2x^2+0x\right)-\left(2x^2-4x\right)=4x$(2x² + 0x) − (2x² − 4x) = 4x, and bring down − 8 to get 4x − 8.

Round 3: $4x/x=4$4x / x = 4. Multiply back: 4(x − 2) = 4x − 8. Subtract: the remainder is 0. So $x^3-8=\left(x-2\right)\left(x^2+2x+4\right)$x³ − 8 = (x − 2)(x² + 2x + 4), and x − 2 is a factor of $x^3-8$x³ − 8. Check by multiplying back: $x^3+2x^2+4x-2x^2-4x-8=x^3-8$x³ + 2x² + 4x − 2x² − 4x − 8 = x³ − 8.

Subtracting a negative term

Divide $2x^3-3x^2+4x-5$2x³ − 3x² + 4x − 5 by x − 1. Round 1: $2x^3/x=2x^2$2x³ / x = 2x², and $2x^2\left(x-1\right)=2x^3-2x^2$2x²(x − 1) = 2x³ − 2x². Subtract: $\left(2x^3-3x^2\right)-\left(2x^3-2x^2\right)=-3x^2+2x^2=-x^2$(2x³ − 3x²) − (2x³ − 2x²) = −3x² + 2x² = −x². Bring down 4x: $-x^2+4x$−x² + 4x.

Round 2: $-x^2/x=-x$−x² / x = −x, and $-x\left(x-1\right)=-x^2+x$−x(x − 1) = −x² + x. Subtract: $\left(-x^2+4x\right)-\left(-x^2+x\right)=4x-x=3x$(−x² + 4x) − (−x² + x) = 4x − x = 3x. Bring down − 5: 3x − 5.

Round 3: $3x/x=3$3x / x = 3, and 3(x − 1) = 3x − 3. Subtract: (3x − 5) − (3x − 3) = −2. The quotient is $2x^2-x+3$2x² − x + 3 and the remainder is −2. Unlike with whole numbers, a remainder can be negative. Check at x = 1, where x − 1 is 0: 2 − 3 + 4 − 5 = −2, the remainder.

Dividing by a quadratic

The loop is the same when the divisor has a higher degree. Divide $x^3+2x^2+3x+4$x³ + 2x² + 3x + 4 by $x^2+1$x² + 1, writing the divisor as $x^2+0x+1$x² + 0x + 1. Round 1: $x^3/x^2=x$x³ / x² = x, and $x\left(x^2+0x+1\right)=x^3+0x^2+x$x(x² + 0x + 1) = x³ + 0x² + x. Subtract from $x^3+2x^2+3x$x³ + 2x² + 3x to get $2x^2+2x$2x² + 2x, and bring down 4: $2x^2+2x+4$2x² + 2x + 4.

Round 2: $2x^2/x^2=2$2x² / x² = 2, and $2\left(x^2+0x+1\right)=2x^2+0x+2$2(x² + 0x + 1) = 2x² + 0x + 2. Subtract: 2x + 2. That has degree 1, lower than the divisor's degree 2, so the loop stops. The quotient is x + 2 and the remainder is 2x + 2.

Check: $\left(x^2+1\right)\left(x+2\right)+2x+2=x^3+2x^2+x+2+2x+2=x^3+2x^2+3x+4$(x² + 1)(x + 2) + 2x + 2 = x³ + 2x² + x + 2 + 2x + 2 = x³ + 2x² + 3x + 4.

The usual mistakes

Subtracting only the first term. In the first example the whole of $x^2+2x$x² + 2x is subtracted, so 5x − 2x = 3x goes forward. Carrying 5x instead gives the wrong quotient, x + 5, and a remainder that does not check.

Copying the divisor into the quotient. The quotient is what the divisor is multiplied by: x + 3, not x + 2.

Taking the last constant as the remainder. In the first example the 7 is the constant before anything is taken away, and 6 is what the quotient uses up; the remainder is 7 − 6 = 1.

Losing a sign when subtracting a negative term. $\left(-x^2+4x\right)-\left(-x^2+x\right)$(−x² + 4x) − (−x² + x) is 3x, because subtracting $-x^2$−x² adds $x^2$x².

Leaving out a missing power. Without $0x^2$0x² and 0x, the terms of $x^3-8$x³ − 8 fall into the wrong places, just as 105 written without its 0 becomes 15.

Stopping too soon or too late. Keep going while what is left has a degree at least that of the divisor, and stop as soon as it is lower.

A look ahead

Writing $\frac{\left(x^2+5x+7\right)}{\left(x+2\right)}$(x² + 5x + 7)/(x + 2) as $x+3+\frac{1}{\left(x+2\right)}$x + 3 + 1/(x + 2) says something about its graph. When x is large, $\frac{1}{\left(x+2\right)}$1/(x + 2) is tiny, so the graph runs very close to the straight line y = x + 3. That line is called a slant asymptote, and it is found by exactly this division.