Factor, remainder, and the cubes
Tonight's idea
Divide a polynomial \(p(x)\) by \((x - a)\), the way you divide 17 by 5 to get 3 remainder 2. You get a quotient \(q(x)\) and a remainder, and because the divisor has degree 1 the remainder is a plain number \(r\): \[ p(x) = (x - a)\,q(x) + r \quad\text{for every } x. \] Put \(x = a\). The first term is zero, leaving \(p(a) = r\). That is the remainder theorem: the remainder on dividing by \((x - a)\) is \(p(a)\). No division required.
The factor theorem follows in one line: \((x - a)\) is a factor exactly when the remainder is zero, so \((x - a)\) divides \(p(x)\) if and only if \(p(a) = 0\). To find the quotient, use synthetic division, shown below.
Two factorisations to know by heart: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2), \qquad a^3 + b^3 = (a + b)(a^2 - ab + b^2). \] The factor theorem predicts the first: \(a = b\) makes \(a^3 - b^3\) zero, so \((a - b)\) must divide it.
Watch one: factor a cubic completely
\(p(x) = x^3 - 6x^2 + 11x - 6\). Try small divisors of the constant term 6. \(p(1) = 1 - 6 + 11 - 6 = 0\), so \((x - 1)\) is a factor. Synthetic division by 1:
| \(a = 1\) | 1 | −6 | 11 | −6 | ← coefficients of \(p\) |
| ↓ | 1 | −5 | 6 | ← \(a \times\) the number below-left | |
| 1 | −5 | 6 | 0 | ← row 1 + row 2 |
Bring the first 1 down. Then repeat: multiply by \(a\), write it under the next coefficient, add. The last entry of the bottom row is the remainder; the others are the quotient, one degree lower: \(x^2 - 5x + 6 = (x - 2)(x - 3)\), remainder 0. So \(p(x) = (x-1)(x-2)(x-3)\). And the remainder on dividing the same \(p(x)\) by \((x - 4)\), without dividing: \(p(4) = 64 - 96 + 44 - 6 = 6\).
Your turn
- Without dividing, find the remainder when \(x^4 - 3x^3 + 2x - 5\) is divided by \((x - 2)\), and then by \((x + 1)\). (Careful: \(x + 1 = x - (-1)\).)
- Factor \(x^3 - 7x + 6\) completely. Find one root among the divisors of 6, then use synthetic division.
- Factor each: \(x^3 - 27\) and \(8x^3 + 125\).
- Polynomial detective. A cubic \(p(x)\) has leading coefficient 1. You are told \(p(2) = 0\), \(p(-1) = 0\) and \(p(0) = 8\). Find \(p(x)\) in expanded form.
Stretch: a factorisation nobody expects
\(x^5 + x + 1\) has no rational roots (try \(\pm1\)), yet it factors. Write it as \(x^5 - x^2 + x^2 + x + 1\), factor \(x^2\) out of the first two terms, and use the difference of cubes on what appears. Then plug in \(x = 2\) to find a factor of 35.
Answers
- \(p(2) = 16 - 24 + 4 - 5 = -9\). \(p(-1) = 1 + 3 - 2 - 5 = -3\).
- \(p(1) = 0\). Synthetic division by 1 gives \(x^2 + x - 6 = (x + 3)(x - 2)\). So \(x^3 - 7x + 6 = (x - 1)(x - 2)(x + 3)\).
- \(x^3 - 27 = (x - 3)(x^2 + 3x + 9)\). \(8x^3 + 125 = (2x + 5)(4x^2 - 10x + 25)\).
- Roots 2 and \(-1\) give factors \((x - 2)(x + 1)\); the third factor is \((x - a)\). \(p(0) = (-2)(1)(-a) = 2a = 8\), so \(a = 4\). \(p(x) = (x-2)(x+1)(x-4) = x^3 - 5x^2 + 2x + 8\).
Stretch. \(x^5 + x + 1 = x^2(x^3 - 1) + (x^2 + x + 1) = x^2(x - 1)(x^2 + x + 1) + (x^2 + x + 1) = (x^2 + x + 1)(x^3 - x^2 + 1)\). At \(x = 2\): \(7 \times 5 = 35\).