Factor and Remainder Theorem

Substitute, and you have the remainder.

Division written as a multiplication

Dividing 17 by 5 gives 3 remainder 2. The same fact can be written without a division sign: 17 = 5 × 3 + 2. The divisor times the quotient, plus the remainder, gives back the number that was divided.

The remainder is always smaller than the divisor. If it were 5 or more, another 5 would fit, and the quotient would be larger.

Polynomials divide the same way

Divide a polynomial f(x) by a bracket x − a and the same sentence holds: f(x) = (x − a) q(x) + r. Here q(x) is the quotient, another polynomial, and r is the remainder.

The remainder must have a lower degree than the divisor, just as 2 is smaller than 5. The divisor x − a has degree 1, so the remainder has degree 0. That means r is a constant: it has no x in it.

Substitute x = a

This identity is true for every value of x, so put x = a into it: f(a) = (a − a) q(a) + r. The bracket is a − a = 0.

Zero times q(a) is zero, whatever q(a) is, so that whole term disappears: f(a) = 0 + r = r. This is the remainder theorem: when f(x) is divided by x − a, the remainder is f(a).

So the remainder can be found by substitution, without dividing at all.

Try it on a number

Divide f(x) = x² + 5 by x − 3. The remainder theorem says the remainder is f(3) = 3² + 5 = 9 + 5 = 14.

Check by long division. x² / x = x, and x(x − 3) = x² − 3x; subtracting from x² + 0x + 5 leaves 3x + 5. Then 3x / x = 3, and 3(x − 3) = 3x − 9; subtracting leaves 5 − (−9) = 14. The quotient is x + 3 and the remainder is 14.

Check by multiplying back: (x − 3)(x + 3) = x² − 9, and x² − 9 + 14 = x² + 5, the polynomial we started with.

A remainder of zero means a factor

If f(a) comes out as 0, the remainder is 0, so f(x) = (x − a) q(x) exactly. Then x − a divides f(x) with nothing left over: it is a factor. This is the factor theorem: x − a is a factor of f(x) exactly when f(a) = 0.

Take f(x) = x³ − 6x² + 11x − 6. At x = 1, f(1) = 1 − 6 + 11 − 6 = 0, so x − 1 is a factor. At x = 4, f(4) = 64 − 96 + 44 − 6 = 6, not 0, so x − 4 is not a factor: dividing by it leaves a remainder of 6.

Dividing by x − 1 leaves x² − 5x + 6 = (x − 2)(x − 3), so f(x) = (x − 1)(x − 2)(x − 3). Check at x = 0: the brackets give (−1) × (−2) × (−3) = −6, the constant term.

Which values to try

Multiply out (x − p)(x − q)(x − s) and the constant term is −pqs. So when the coefficients are whole numbers and the x³ coefficient is 1, a whole-number root must divide the constant term. For x³ − 6x² + 11x − 6, only ±1, ±2, ±3 and ±6 are worth testing.

Roots on the graph

A factor x − a makes f(a) = 0, so the graph of y = f(x) meets the x-axis at x = a. For x³ − 6x² + 11x − 6 that happens at x = 1, 2 and 3.

Where f(a) is not zero, f(a) is the height of the curve at x = a, and that height is the remainder. At x = 4 the curve is 6 above the x-axis, and dividing by x − 4 leaves 6.

xy

The curve y = x³ − 6x² + 11x − 6. It crosses the x-axis at x = 1, 2 and 3, the roots that give the factors x − 1, x − 2 and x − 3. At x = 4 the dot is 6 above the axis: f(4) = 6, the remainder on dividing by x − 4.

Two more uses

A divisor that does not start with x works the same way: the remainder is the value of f at the x that makes the divisor zero. For 2x − 1 that x is ½. With f(x) = 2x³ + x² − 5x + 2, f(½) = 2 × (½)³ + (½)² − 5 × ½ + 2 = 0.25 + 0.25 − 2.5 + 2 = 0, so 2x − 1 is a factor. Indeed 2x³ + x² − 5x + 2 = (2x − 1)(x² + x − 2) = (2x − 1)(x + 2)(x − 1).

A known remainder can find a missing coefficient. If x³ + kx − 4 leaves a remainder of 5 when divided by x − 3, then f(3) = 27 + 3k − 4 = 5. So 3k = −18 and k = −6. Check: 27 − 18 − 4 = 5.

The usual mistakes

Substituting the wrong sign. Dividing by x − 3 means substituting x = 3, and x + 3 means x = −3. The value is the one that makes the bracket zero.

Taking the constant term as the remainder. For x² + 5 divided by x − 3, the 5 is f(0); the remainder is f(3) = 14.

Skipping the square. f(3) for x² + 5 is 3 × 3 + 5 = 14, not 3 + 5 = 8.

Testing a coefficient instead of a root. For x² − 7x + 12, the middle coefficient 7 is the sum of the roots 3 and 4; f(7) = 49 − 49 + 12 = 12, so x − 7 is not a factor, but f(3) = 9 − 21 + 12 = 0, so x − 3 is.

A trail over a hump

In the application below, a trail has height x³ + ax² + bx + 6. A point level with the car park is a root, so it gives a factor and one equation. A known height at another point is a remainder, and gives the second equation.

Worked example: A Mountain-Bike Trail over a Hump and Through a Dip: Two Missing Coefficients from the Factor and Remainder Theorems

Question A mountain-bike trail runs over a hump and down through a dip. Its height above the level of the car park is y = x3 + ax2 + bx + 6 meters, where x is the distance along the trail in tens of meters and 0 ≤ x ≤ 3.5. (a) The trail is level with the car park at x = 1, and at x = 2 it is 4 m below the car park. Use the factor theorem and the remainder theorem to find a and b. (b) Find the other point on the trail where it is level with the car park, and say why the third root of the cubic is not on the trail.

  1. 1.The trail is level with the car park at x = 1, so y = 0 there, and by the factor theorem (x − 1) is a factor: 1 + a + b + 6 = 0, which gives a + b = −7.

    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)y = 0 at x = 1: 1 + a + b + 6 = 0a + b = −7
    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)y = 0 at x = 1: 1 + a + b + 6 = 0a + b = −7
    The trail is level with the car park at x = 1, so (x − 1) is a factor: 1 + a + b + 6 = 0, and a + b = −7.

  2. 2.By the remainder theorem, the remainder when the cubic is divided by (x − 2) is its value at x = 2, which is −4 because the trail is below the car park: 8 + 4a + 2b + 6 = −4, so 4a + 2b = −18 and 2a + b = −9.

    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)remainder on dividing by (x − 2) = y at x = 28 + 4a + 2b + 6 = −4, so 2a + b = −9
    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)remainder on dividing by (x − 2) = y at x = 28 + 4a + 2b + 6 = −4, so 2a + b = −9
    At x = 2 the height is −4, the remainder on dividing by (x − 2): 8 + 4a + 2b + 6 = −4, so 2a + b = −9.

  3. 3.Subtract the first equation from the second: a = −2, and then b = −7 − (−2) = −5. (a) a = −2 and b = −5, so y = x3 − 2x2 − 5x + 6.

    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)subtract: a = −2 and b = −5y = x3− 2x2− 5x + 6
    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)subtract: a = −2 and b = −5y = x3− 2x2− 5x + 6
    (a) Subtracting gives a = −2 and b = −5. The curve y = x3 − 2x2 − 5x + 6 passes through both points.

  4. 4.Divide by the factor (x − 1): x3 − 2x2 − 5x + 6 = (x − 1)(x2 − x − 6) = (x − 1)(x − 3)(x + 2).

    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)(x − 1)(x2− x − 6)= (x − 1)(x − 3)(x + 2)
    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)(x − 1)(x2− x − 6)= (x − 1)(x − 3)(x + 2)
    Divide by (x − 1): x3 − 2x2 − 5x + 6 = (x − 1)(x2 − x − 6) = (x − 1)(x − 3)(x + 2).

  5. 5.(b) The trail is level with the car park again at x = 3, which is 30 m along. The root x = −2 lies before the start of the trail, outside 0 ≤ x ≤ 3.5, so it is rejected. Check: at x = 3, 27 − 18 − 15 + 6 = 0.

    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)(3, 0)x = −2level again at x = 3: 30 m alongx = −2 is before the start: rejected
    −8−4048−2−10123tens of meters along the trail, xheight (m), y(1, 0)(2, −4)(3, 0)x = −2level again at x = 3: 30 m alongx = −2 is before the start: rejected
    (b) The trail is level with the car park again at x = 3, 30 m along. The root x = −2 is on the dashed part of the curve, before the trail starts.

Answer: (a) a = −2 and b = −5, so y = x3 − 2x2 − 5x + 6; (b) at x = 3, 30 m along the trail; the root x = −2 lies outside 0 ≤ x ≤ 3.5

Common mistakes

  • Finding the remainder on dividing by (x − 2) from the value at x = −2. The remainder on dividing by (x − k) is the value at x = k, here x = 2.
  • Setting the height at x = 2 to 4. The trail is 4 m below the car park there, so its height above the car park is −4.

More polynomials and the binomial theorem problems, worked step by step →

Practice Factor and Remainder Theorem in the app