Dividing polynomials and finding roots
The remainder theorem, long division and Horner's synthetic division, and the roots of a cubic: guess an integer root, divide, solve the quadratic.
How to solve it
Polynomials divide with a remainder, like numbers. The main consequence is the remainder theorem: the remainder of $P(x)$ divided by $x - a$ is $P(a)$. So if $P(a) = 0$, the polynomial is divisible by $x - a$, and the degree of the equation can be lowered.
Step by step
- Divisibility by $x - a$: put $a$ into the polynomial; it divides if the result is zero. For $x + 3$ put $a = -3$.
- Dividing by $x - a$ is easiest with Horner's scheme: bring down the leading coefficient, multiply by $a$, add to the next one, and so on. The last number is the remainder.
- Divide by a quadratic with long division: divide the leading terms, multiply back, subtract.
- Roots: the integer roots of a polynomial with integer coefficients are among the divisors of the constant term. Having found one, divide by $x - a$ and solve the quadratic that is left.
Common mistakes
- Putting a number with the wrong sign into the theorem: the divisor $x + 3$ is $x - (-3)$, so use $-3$.
- Skipping zero coefficients in Horner's scheme: $x^3 - 11x - 20$ has the coefficients $1, 0, -11, -20$.
- Sign errors when subtracting in long division: the whole product is subtracted, every term of it.
- Looking for integer roots among arbitrary numbers. The candidates are only the divisors of the constant term, with both signs.