Topic 9 Long division and factorisation

9.1 Polynomials and long division

A polynomial is an expression of the form \[p(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0, \] where \(a_0,\dots,a_n\) are constants, also called coefficients. Its degree (or order) is the highest occurring power of \(x\); in other words if \(a_n\neq 0\), then the degree of \(p(x)\) above is \(n\).

The low–degree polynomials have special names:

example polynomial its degree also called
\(x-7\) 1 linear polynomial
\(3x^2+2\) 2 quadratic polynomial
\(4x^3+5x^2-3x\) 3 cubic polynomial
\(-x^4+x^2-2\) 4 quartic polynomial

Recall long division of numbers: \[ \begin{array}{rl} 205&\text{ r. }18 \\[-3pt] 21 \overline{){4323}}\kern-.2ex \\[-3pt] \underline{-42}\phantom{m}\\[-3pt] 123 \\[-3pt] \underline{-105}\\[-3pt] 18 \end{array} \quad\quad\quad\text{or}\quad\quad\quad \begin{array}{rl} 4323& \div \,\,21 = 205\text{ r. }18 \\ \underline{-42}\phantom{m}\\[-3pt] 123 \\[-3pt] \underline{-105}\\[-3pt] 18 \end{array} \] We can write this as \[ \frac{4323}{21} = 205+\frac{18}{21}\quad\quad\text{or}\quad\quad 4323 = 205\cdot 21 + 18. \] In general: \[ \frac{\text{dividend}}{\text{divisor}} = \text{quotient} + \frac{\text{remainder}}{\text{divisor}} \] or \[ \text{dividend} = \text{quotient}\cdot\text{divisor} + \text{remainder}. \] Long division of polynomials works very similarly: \[ \begin{array}{rl} (x^4-x^3+x^2-3x+5)&\div\,\,\, (x-1) = x^3+x-2\quad\text{r. }3\\ \underline{-(x^4 -x^3)}\phantom{+x^2-3x+5)}&\\ 0+x^2-3x\phantom{+5x)}\\ \underline{-(x^2-x)}\phantom{+5x)}\\ -2x+5\phantom{m}\\ \underline{-(-2x+2)}\phantom{)}\\ 3\phantom{m} \end{array} \] If we write \(p(x):=x^4-x^3+x^2-3x+5\) (for the dividend) and \(q(x):=x^3+x-2\) (for the quotient), we can write \[ p(x) = q(x)\cdot(x-1) + 3. \] This holds for any \(x\), so in particular for \(x=1\) we get \(p(1) = q(1)\cdot0+3=3\).

Theorem 9.1 (Remainder Theorem) The remainder left when a polynomial \(p(x)\) is divided by \(x-a\) is equal to \(p(a)\). In particular, \(x-a\) is a factor of \(p(x)\) if an only if \(p(a)=0\).

Example 9.1 Determine the remainder when \(p(x)=5x^3+2x^2-6\) is divided by \(x-2\).

Solution. The remainder is \(p(2) = 5\cdot8+2\cdot4-6=42\).

Example 9.2 Divide \(3x^3-16x^2+15x+18\) by \(x-3\): \[ \begin{array}{rl} (3x^3-16x^2+15x+18) &\div \,\,\,(x-3) = 3x^2-7x-6\\ \underline{-(3x^3-9x^2)}\phantom{+15x+18x}\\ -7x^2 + 15x\phantom{+18x)}\\ \underline{-(-7x^2+21x)}\phantom{+18o}\\ -6x+18\phantom{x}\\ \underline{-(-6x+18)}\\ 0\phantom{x} \end{array} \] Thus \(3x^3-16x^2+15x+18 = (3x^2-7x-6)(x-3)\).
Using a quadratic formula or by inspection, we can further factorise the quadratic as \(3x^2-7x-6 = (x-3)(3x+2)\).
Altogether, we obtain complete factorisation \[3x^3-16x^2+15x+18=(x-3)^2(3x+2).\]

9.2 Obtaining the factorisation of a polynomial

We start with a polynomial \(p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\). We wish to factorise it as much as possible, ideally into linear factors.

(Note that this is a heuristic algorithm, i.e. it may not always work. The point is that getting a factorisation of a polynomial in the end amounts to finding its roots — and Abel’s impossibility theorem asserts that there is no algebraic formula that would yield the solutions of an equation \(p(x)=0\) when degree of the polynomial \(p(x)\) is 5 or higher.)

Step 1. (“Guess a root.”) Calculate \(p(b)\) for divisors \(b\) of \(a_0\). When we find \(b\) such that \(p(b)=0\), we know that \(x-b\) is a factor of \(p(x)\), and we perform

Step 2. (“Divide.”) Calculate \(q(x) := p(x)\div(x-b)\) using long division.

Repeat. Apply Steps 1. and 2. to the “new” polynomial \(q(x)\) (in place of \(p(x)\).\ Note that \(q(x)\) has degree one less than the degree of \(p(x)\), so eventually we end up with a linear polynomial, and so we are done.

Example 9.3 Factorise \(p(x)=x^4-x^3+x^2-3x+2\).

Solution. \(p(1)=1-1+1-3+2=0\) \(\implies\) \(x-1\) is a factor of \(p(x)\). Next, we perform long division: \[ \begin{array}{rl} (x^4-x^3+x^2-3x+2) & \div \,\,\,(x-1) = x^3+x-2 =: q(x)\\ \underline{-(x^4-x^3)}\phantom{+x^2-3x+2x}\\ 0 + x^2-3x\phantom{+2m}\\ \underline{-(x^2-x)}\phantom{+2x}\\ -2x+2\phantom{x}\\ \underline{-(-2x+2)}\\ 0\phantom{x} \end{array} \] Next, we now look for a root of \(q(x)\).
\(q(1) = 1+1-2=0\) \(\implies\) \(x-1\) is a factor of \(q(x)\). We continue by performing long division: \[ \begin{array}{rl} (x^3\phantom{-xm}+x-2) & \div \,\,\,(x-1) = x^2+x+1\\ \underline{-(x^3-x^2)}\phantom{x+2xx}\\ 0 + x^2+x\phantom{+2m}\\ \underline{-(x^2-x)}\phantom{+2x}\\ 2x-2\phantom{x}\\ \underline{-(2x-2)}\\ 0\phantom{x} \end{array} \] We now try to continue by further factorising \(x^2+x+1\). However the discriminant of this quadratic is \(1-4\cdot2=-7<0\), so it can not be further factorised.
Answer: \(p(x)=(x-1)^2(x^2+x+1)\).

Example 9.4 \[ \begin{array}{rl} (x^3+5x^2-7x+3) &\div\,\,\,(x^2+2x+1) = x+3\quad\text{ rem. }-14x\\ \underline{-(x^3+2x^2+\phantom{7}x)}\phantom{+3)}\\ 3x^2-8x+3\phantom{)}\\ \underline{-(3x^2+6x+3)}\\ -14x\phantom{+3m} \end{array} \]

Example 9.5 The polynomial \(p(x)=6x^3-23x^2+ax+b\) leaves remainder \(11\) when divided by \(x-3\) and remainder \(-21\) when divided by \(x+1\).

  1. Show that \(a\) and \(b\) satisfy the equations \(3a+b=56\), \(-a+b=8\).
  2. Solve the simultaneous equations above.
  3. Show that \(x-2\) is a factor of \(p(x)\).
  4. Factorise \(p(x)\).
  5. Determine the roots of \(p(x)\).

  1. Deriving the equations:

    • \(p(x)\) leaves a remainder of \(11\) when divided by \(x-3\)
      \(\iff\) \(p(3)=11\)  (by Remainder Theorem)
      \(\iff\) \(6\cdot 27-23\cdot 9+3a+b = 11\)
      \(\iff\) \(3a+b=56\)       (A)
    • \(p(x)\) leaves a remainder of \(-21\) when divided by \(x+1\)
      \(\iff\) \(p(-1)=-21\)  (by Remainder Theorem)
      \(\iff\) \(-6-23-a+b=-21\)
      \(\iff\) \(-a+b=8\)       (B)

  2. From (A): \(b=56-3a\).
    Substitute into (B): \(-a+56-3a = 8\)
    \(\iff\) \(4a=48\)
    \(\iff\) \(a=12\). Substitute back into (B): \(-12+b=8\)
    \(\iff\) \(b=20\).

  3. From (ii): \(p(x)=6x^3-23x^2+12x+20\).
    \(p(2)=6\cdot8 - 23\cdot 4+12\cdot2+20=48-92+24+20=0\), so \(x-2\) is a factor of \(p(x)\) by Remainder Theorem.

  4. We start by long division by the factor we already know from (iii). \[ \begin{array}{rl} (6x^3-23x^2+12x+20) &\div\,\,\,(x-2) = 6x^2-11x-10\\ \underline{-(6x^3-12x^2)}\phantom{+12x+20)}\\ -11x^2+12x\phantom{+20m}\\ \underline{-(-11x^2+22x)}\phantom{+20)}\\ -10x+20\phantom{)}\\ \underline{-(-10x+20)}\\ 0\phantom{)} \end{array} \] By inspection: \(6x^2-11x-10=(2x-5)(3x+2)\). So altogether \(p(x)=(x-2)(2x-5)(3x+2)\).

  5. From (iv): the roots of \(p(x)\) are \(x=2\), \(x=\frac{5}{2}\) and \(x=-\frac{2}{3}\).