A polynomial f(x) is exactly divisible by d(x) when the remainder of the division is identically zero. This section shows how to find the conditions that the coefficients of f(x) must satisfy for such exact division to occur, and why those conditions are equivalent to d(x) being a factor of f(x).
Quick Reference
| Situation | What to do |
|---|---|
| Divide f(x) by d(x), deg(d) = m | Perform long division; remainder r(x) has degree at most m − 1 |
| Exact divisibility condition | Set all m coefficients of r(x) equal to zero |
| Number of conditions | Always equal to deg(d) = m |
| What exact divisibility means | d(x) is a factor of f(x); equivalently, every root of d(x) is a root of f(x) |
The Number of Conditions
Let f(x) and d(x) be polynomials with literal coefficients, and suppose deg(d) = m. By the Division Algorithm, there exist unique polynomials q(x) and r(x) such that
$f(x) = d(x)\,q(x) + r(x),$where r(x) = 0 or deg(r) < m. For the division to be exact, r(x) must be identically zero, meaning every one of its (at most m) coefficients must be zero. Since each of these coefficients depends on the coefficients of f(x) and d(x), we obtain a system of conditions on those coefficients.
Exact Divisibility Theorem. In order that a polynomial f(x) be exactly divisible by a polynomial d(x) of degree m (that is, m = deg(d)), the coefficients of f(x) and d(x) must satisfy exactly m conditions (obtained by setting each coefficient of the remainder equal to zero).
- When the division is exact (that is, when r(x) = 0), the Division Algorithm reduces to f(x) = d(x) q(x), which means d(x) is a factor of f(x). Finding conditions for exact divisibility is therefore equivalent to finding conditions under which d(x) divides f(x) without remainder.
Worked Examples
Example 1. For what values of a and b is x3 + 3x2 + bx + 2 exactly divisible by x2 + ax + 1?
Solution
Let f(x) = x3 + 3x2 + bx + 2 and d(x) = x2 + ax + 1. Since deg(d) = 2, there will be exactly 2 conditions on a and b, which is precisely what we need to determine both unknowns. We perform long division of f(x) by d(x): $\begin{array}{ll} & \quad x+(3-a) \\ x^2+ax+1 & \overline{)\ x^3+3x^2+bx+2} \\ & \underline{\ x^3+ax^2+x\phantom{{}+bx+2\ }} \\ & \phantom{x^3+{}}(3-a)x^2+(b-1)x+2 \\ & \underline{\ (3-a)x^2+a(3-a)x+(3-a)} \\ & \phantom{(3-a)x^2+{}}\bigl(b-1-3a+a^2\bigr)x+(a-1) \end{array}$ The quotient is q(x) = x + (3 − a) and the remainder is: $r(x) = \bigl(b-1-3a+a^2\bigr)x+(a-1).$ For exact divisibility, both coefficients of r(x) must be zero: $b-1-3a+a^2 = 0, \qquad a-1 = 0.$ Solving sequentially: from the second equation, a = 1; substituting into the first: $b-1-3+1 = 0 \implies b = 3.$ $\boxed{a = 1, \quad b = 3.}$ Verification. With a = 1 and b = 3, we have f(x) = x3 + 3x2 + 3x + 2 and d(x) = x2 + x + 1. Confirming: $(x^2+x+1)(x+2) = x^3+2x^2+x^2+2x+x+2 = x^3+3x^2+3x+2. \checkmark$Example 2. Determine l and m so that 2x3 + 3x2 + lx + m may be exactly divisible by x2 + x − 6.
Solution
Let f(x) = 2x3 + 3x2 + lx + m and d(x) = x2 + x − 6. Since deg(d) = 2, exact divisibility imposes 2 conditions on l and m, exactly enough to determine both unknowns. $\begin{array}{ll} & \quad 2x+1 \\ x^2+x-6 & \overline{)\ 2x^3+3x^2+lx+m} \\ & \underline{\ 2x^3+2x^2-12x\phantom{{}+m\ \ }} \\ & \phantom{2x^3+{}}x^2+(l+12)x+m \\ & \underline{\ x^2+x-6\phantom{{}+(l+12)x}} \\ & \phantom{x^2+{}}(l+11)x+(m+6) \end{array}$ The quotient is q(x) = 2x + 1 and the remainder is: $r(x) = (l+11)x+(m+6).$ For exact divisibility, both coefficients of r(x) must be zero: $l+11 = 0 \implies l = -11,$ $m+6 = 0 \implies m = -6.$ $\boxed{l = -11, \quad m = -6.}$ Verification. With l = −11 and m = −6: $f(x) = 2x^3 + 3x^2 - 11x - 6.$ Confirming f(x) = d(x) q(x): $ \begin{aligned} (x^2+x-6)(2x+1) &= 2x^3+x^2+2x^2+x-12x-6 \\ &= 2x^3+3x^2-11x-6. \checkmark \end{aligned} $
