3.4 Komplexe Polynome

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Zeile 28: Zeile 28:
An expression of the form
An expression of the form
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{{Displayed math||<math>a_nx^n+a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x+a_0</math>}}
+
{{Abgesetzte Formel||<math>a_nx^n+a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x+a_0</math>}}
where <math>n</math> is an integer, is called a ''polynomial'' of degree <math>n</math> in an unknown variable <math>x</math>. The number <math>a_1</math> is called the coefficient of <math>x</math>, <math>a_2</math> the coefficient of <math>x^2</math>, etc. The constant <math>a_0</math> is called the ''constant term''.
where <math>n</math> is an integer, is called a ''polynomial'' of degree <math>n</math> in an unknown variable <math>x</math>. The number <math>a_1</math> is called the coefficient of <math>x</math>, <math>a_2</math> the coefficient of <math>x^2</math>, etc. The constant <math>a_0</math> is called the ''constant term''.
Zeile 42: Zeile 42:
Compare the following integer written using a base 10,
Compare the following integer written using a base 10,
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{{Displayed math||<math>1353= 1\times 10^3 + 3\times 10^2 + 5\times 10 + 3</math>}}
+
{{Abgesetzte Formel||<math>1353= 1\times 10^3 + 3\times 10^2 + 5\times 10 + 3</math>}}
with the polynomial in <math>x</math>
with the polynomial in <math>x</math>
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{{Displayed math||<math>x^3 + 3x^2 + 5x + 3 = 1\times x^3 + 3\times x^2 + 5\times x + 3</math>}}
+
{{Abgesetzte Formel||<math>x^3 + 3x^2 + 5x + 3 = 1\times x^3 + 3\times x^2 + 5\times x + 3</math>}}
and then the following divisions,
and then the following divisions,
Zeile 60: Zeile 60:
As the above example showed, polynomials can be divided very like integers. Polynomial division, like integer division, is usually not exact. If, for example, <math>37</math> is divided by <math>5</math>, one gets
As the above example showed, polynomials can be divided very like integers. Polynomial division, like integer division, is usually not exact. If, for example, <math>37</math> is divided by <math>5</math>, one gets
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{{Displayed math||<math>\frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}</math>}}
The calculation can also be written as <math>\ 37= 7\times 5+2\,</math>. The number 7 is called the ''quotient'' and the number 2 the ''remainder''. One says that dividing 37 by 5 gives the quotient 7 and the remainder 2.
The calculation can also be written as <math>\ 37= 7\times 5+2\,</math>. The number 7 is called the ''quotient'' and the number 2 the ''remainder''. One says that dividing 37 by 5 gives the quotient 7 and the remainder 2.
Zeile 67: Zeile 67:
Similarly, if <math>p(x)</math> and <math>q(x)</math> are polynomials, one can divide <math>p(x)</math> by <math>q(x)</math> and unambiguously determine polynomials <math>k(x)</math> and <math>r(x)</math> such that
Similarly, if <math>p(x)</math> and <math>q(x)</math> are polynomials, one can divide <math>p(x)</math> by <math>q(x)</math> and unambiguously determine polynomials <math>k(x)</math> and <math>r(x)</math> such that
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{{Displayed math||<math>\frac{p(x)}{q(x)} = k(x)+ \frac{r(x)}{q(x)}\,\mbox{,}</math>}}
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{{Abgesetzte Formel||<math>\frac{p(x)}{q(x)} = k(x)+ \frac{r(x)}{q(x)}\,\mbox{,}</math>}}
or <math>\ p(x)= k(x)\, q(x)+r(x)\,</math>. One here says that polynomial division has resulted in a quotient <math>k(x)</math> and remainder <math>r(x)</math>.
or <math>\ p(x)= k(x)\, q(x)+r(x)\,</math>. One here says that polynomial division has resulted in a quotient <math>k(x)</math> and remainder <math>r(x)</math>.
Zeile 74: Zeile 74:
Clearly, a division is exact if the remainder is zero. For polynomials this is expressed as follows: If <math>r(x)=0</math> then <math>p(x)</math> is divisible by <math>q(x)</math>, or, <math>q(x)</math> is a ''divisor'' of <math>p(x)</math>. One writes
Clearly, a division is exact if the remainder is zero. For polynomials this is expressed as follows: If <math>r(x)=0</math> then <math>p(x)</math> is divisible by <math>q(x)</math>, or, <math>q(x)</math> is a ''divisor'' of <math>p(x)</math>. One writes
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{{Displayed math||<math>\frac{p(x)}{q(x)} = k(x)\,\mbox{,}</math>}}
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{{Abgesetzte Formel||<math>\frac{p(x)}{q(x)} = k(x)\,\mbox{,}</math>}}
or <math>\ p(x) = k(x)\, q(x)\,</math>.
or <math>\ p(x) = k(x)\, q(x)\,</math>.
Zeile 92: Zeile 92:
The first step is that we ''add and subtract'' an appropriate <math>x^2</math>-term in the numerator
The first step is that we ''add and subtract'' an appropriate <math>x^2</math>-term in the numerator
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{{Displayed math||<math>\frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}</math>}}
The reason why we do this is that the sub-expression <math>x^3+2x^2</math> can be written as <math>x^2(x+2)</math> and cancellation with the denominator can be done,
The reason why we do this is that the sub-expression <math>x^3+2x^2</math> can be written as <math>x^2(x+2)</math> and cancellation with the denominator can be done,
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{{Displayed math||<math>\frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}</math>}}
Then we add and subtract an appropriate <math>x</math>-term so that the leading <math>x^2</math>-term in the numerator can be cancelled,
Then we add and subtract an appropriate <math>x</math>-term so that the leading <math>x^2</math>-term in the numerator can be cancelled,
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{{Displayed math||<math>\begin{align*} x^2+\frac{-x^2-2x+2x-x+4}{x+2} &= x^2+\frac{-x(x+2)+2x-x+4}{x+2}\\ &=x^2-x+\frac{x+4}{x+2}\,\mbox{.}\end{align*}</math>}}
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{{Abgesetzte Formel||<math>\begin{align*} x^2+\frac{-x^2-2x+2x-x+4}{x+2} &= x^2+\frac{-x(x+2)+2x-x+4}{x+2}\\ &=x^2-x+\frac{x+4}{x+2}\,\mbox{.}\end{align*}</math>}}
The last step is that we add and subtract a constant
The last step is that we add and subtract a constant
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{{Displayed math||<math>x^2-x+\frac{x+4}{x+2}=x^2-x+\frac{x+2-2+4}{x+2} = x^2-x+1+\frac{2}{x+2}\,\mbox{.}</math>}}
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{{Abgesetzte Formel||<math>x^2-x+\frac{x+4}{x+2}=x^2-x+\frac{x+2-2+4}{x+2} = x^2-x+1+\frac{2}{x+2}\,\mbox{.}</math>}}
Thus ending up with
Thus ending up with
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{{Displayed math||<math>\frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}</math>}}
The quotient is <math>x^2 -x + 1</math> and the remainder is <math>2</math>. Since the remainder is not zero, division is not exact, that is, <math>q(x)= x+2</math> is not a ''divisor'' of <math>p(x)=x^3 + x^2 -x +4</math>.
The quotient is <math>x^2 -x + 1</math> and the remainder is <math>2</math>. Since the remainder is not zero, division is not exact, that is, <math>q(x)= x+2</math> is not a ''divisor'' of <math>p(x)=x^3 + x^2 -x +4</math>.
Zeile 119: Zeile 119:
If <math>q(x)</math> is a divisor of <math>p(x)</math> then <math>p(x)=k(x)\, q(x)</math>. We have thus ''factorised'' <math>p(x)</math> . One says that <math>q(x)</math> is a factor of <math>p(x)</math>. Especially, if a polynomial of first degree <math>(x-a)</math> is a divisor of <math>p(x)</math> then <math>(x-a)</math> is a factor of <math>p(x)</math> , i.e.
If <math>q(x)</math> is a divisor of <math>p(x)</math> then <math>p(x)=k(x)\, q(x)</math>. We have thus ''factorised'' <math>p(x)</math> . One says that <math>q(x)</math> is a factor of <math>p(x)</math>. Especially, if a polynomial of first degree <math>(x-a)</math> is a divisor of <math>p(x)</math> then <math>(x-a)</math> is a factor of <math>p(x)</math> , i.e.
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{{Displayed math||<math>p(x)= q(x)\, (x-a)\,\mbox{.}</math>}}
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{{Abgesetzte Formel||<math>p(x)= q(x)\, (x-a)\,\mbox{.}</math>}}
Since <math>\ p(a)=q(a)\, (a-a)= q(a)\times 0 = 0\ </math> this means that <math>x=a</math> is a zero of <math>p(x)</math>. This is exactly the content of the so-called ''factor theorem''.
Since <math>\ p(a)=q(a)\, (a-a)= q(a)\times 0 = 0\ </math> this means that <math>x=a</math> is a zero of <math>p(x)</math>. This is exactly the content of the so-called ''factor theorem''.
Zeile 137: Zeile 137:
The polynomial <math>p(x) = x^2-6x+8</math> can be factorised as
The polynomial <math>p(x) = x^2-6x+8</math> can be factorised as
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{{Displayed math||<math>x^2-6x+8 = (x-2)(x-4)</math>}}
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{{Abgesetzte Formel||<math>x^2-6x+8 = (x-2)(x-4)</math>}}
and has therefore zeros at <math>x=2</math> and <math>x=4</math> (and no others). It is precisely these that are obtained if one solves the equation <math>\ x^2-6x+8 = 0\,</math>.
and has therefore zeros at <math>x=2</math> and <math>x=4</math> (and no others). It is precisely these that are obtained if one solves the equation <math>\ x^2-6x+8 = 0\,</math>.
Zeile 153: Zeile 153:
By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation <math>\ x^2-3x-10=0\ </math> has the solutions
By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation <math>\ x^2-3x-10=0\ </math> has the solutions
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{{Displayed math||<math>x= \frac{3}{2} \pm \sqrt{\Bigl(\frac{3}{2}\Bigr)^2 - (-10)} = \frac{3}{2} \pm \frac{7}{2}\,\mbox{,}</math>}}
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{{Abgesetzte Formel||<math>x= \frac{3}{2} \pm \sqrt{\Bigl(\frac{3}{2}\Bigr)^2 - (-10)} = \frac{3}{2} \pm \frac{7}{2}\,\mbox{,}</math>}}
i.e. <math>x=-2</math> and <math>x=5</math>. This means that <math>\ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,</math>.
i.e. <math>x=-2</math> and <math>x=5</math>. This means that <math>\ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,</math>.
Zeile 163: Zeile 163:
This polynomial has a repeated root
This polynomial has a repeated root
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{{Displayed math||<math>x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3</math>}}
+
{{Abgesetzte Formel||<math>x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3</math>}}
and thus <math>\ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,</math>.
and thus <math>\ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,</math>.
Zeile 173: Zeile 173:
In this case, the polynomial has two complex roots
In this case, the polynomial has two complex roots
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{{Displayed math||<math>x= 2 \pm \sqrt{2^2 -5} = 2\pm \sqrt{-1} = 2\pm i</math>}}
+
{{Abgesetzte Formel||<math>x= 2 \pm \sqrt{2^2 -5} = 2\pm \sqrt{-1} = 2\pm i</math>}}
and when factorised will be <math>\ (x-(2-i))(x-(2+i))\,</math>.
and when factorised will be <math>\ (x-(2-i))(x-(2+i))\,</math>.
Zeile 191: Zeile 191:
According to the factor theorem, the polynomial must have factors <math>(x-1)</math>, <math>(x+1)</math> and <math>(x-3)</math>. Multiplying these factors, we get a cubic polynomial
According to the factor theorem, the polynomial must have factors <math>(x-1)</math>, <math>(x+1)</math> and <math>(x-3)</math>. Multiplying these factors, we get a cubic polynomial
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{{Displayed math||<math>(x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}</math>}}
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{{Abgesetzte Formel||<math>(x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}</math>}}
</div>
</div>
Zeile 224: Zeile 224:
We have
We have
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{{Displayed math||<math>\begin{align*} p(i) &= i^4- 4i^3 +6i^2-4i+5 = 1+4i-6-4i+5=0\,\mbox{,}\\ p(2-i) &= (2-i)^4 -4(2-i)^3 + 6(2-i)^2 - 4(2-i) +5\,\mbox{.}\end{align*}</math>}}
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{{Abgesetzte Formel||<math>\begin{align*} p(i) &= i^4- 4i^3 +6i^2-4i+5 = 1+4i-6-4i+5=0\,\mbox{,}\\ p(2-i) &= (2-i)^4 -4(2-i)^3 + 6(2-i)^2 - 4(2-i) +5\,\mbox{.}\end{align*}</math>}}
In order to calculate the last term, we need to determine
In order to calculate the last term, we need to determine
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{{Displayed math||<math>\begin{align*} (2-i)^2 &= 4-4i+i^2 = 3-4i\,\mbox{,}\\ (2-i)^3 &= (3-4i)(2-i) = 6-3i-8i+4i^2 = 2-11i\,\mbox{,}\\ (2-i)^4 &= (2-11i)(2-i) = 4-2i-22i+11i^2= -7-24i\,\mbox{.}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*} (2-i)^2 &= 4-4i+i^2 = 3-4i\,\mbox{,}\\ (2-i)^3 &= (3-4i)(2-i) = 6-3i-8i+4i^2 = 2-11i\,\mbox{,}\\ (2-i)^4 &= (2-11i)(2-i) = 4-2i-22i+11i^2= -7-24i\,\mbox{.}\end{align*}</math>}}
This gives that
This gives that
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{{Displayed math||<math>\begin{align*} p(2-i) &= -7-24i-4(2-11i)+6(3-4i) -4(2-i) +5\\ &= -7-24i-8+44i+18-24i-8+4i+5=0\,\mbox{,}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*} p(2-i) &= -7-24i-4(2-11i)+6(3-4i) -4(2-i) +5\\ &= -7-24i-8+44i+18-24i-8+4i+5=0\,\mbox{,}\end{align*}</math>}}
which proves that <math>i</math> and <math>2-i</math> are zeros of this polynomial.
which proves that <math>i</math> and <math>2-i</math> are zeros of this polynomial.
Zeile 253: Zeile 253:
We have that <math>\ p(1)= 1^3 + 1^2 -2 = 0\ </math> which shows that <math>x=1</math> is a zero of the polynomial. According to the factor theorem, this means that <math>x-1</math> is a factor of <math>p(x)</math>, i.e. <math>p(x)</math> is divisible by <math>x-1</math>. We therefore divide the polynomial with <math>x-1</math> to get the remaining factors after <math>x-1</math> is factored out of the polynomial
We have that <math>\ p(1)= 1^3 + 1^2 -2 = 0\ </math> which shows that <math>x=1</math> is a zero of the polynomial. According to the factor theorem, this means that <math>x-1</math> is a factor of <math>p(x)</math>, i.e. <math>p(x)</math> is divisible by <math>x-1</math>. We therefore divide the polynomial with <math>x-1</math> to get the remaining factors after <math>x-1</math> is factored out of the polynomial
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{{Displayed math||<math>\begin{align*} \frac{x^3+x^2-2}{x-1} &= \frac{x^2(x-1)+2x^2-2}{x-1} = x^2 + \frac{2x^2-2}{x-1} = x^2 + \frac{2x(x-1) +2x -2}{x-1}\\[4pt] &= x^2 + 2x + \frac{2x-2}{x-1} = x^2 + 2x + \frac{2(x-1)}{x-1} = x^2 + 2x + 2\,\mbox{.}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*} \frac{x^3+x^2-2}{x-1} &= \frac{x^2(x-1)+2x^2-2}{x-1} = x^2 + \frac{2x^2-2}{x-1} = x^2 + \frac{2x(x-1) +2x -2}{x-1}\\[4pt] &= x^2 + 2x + \frac{2x-2}{x-1} = x^2 + 2x + \frac{2(x-1)}{x-1} = x^2 + 2x + 2\,\mbox{.}\end{align*}</math>}}
So we have <math>\ p(x)= (x-1)(x^2+2x+2)\,</math> which is the first part of the problem.
So we have <math>\ p(x)= (x-1)(x^2+2x+2)\,</math> which is the first part of the problem.
Zeile 260: Zeile 260:
It now remains to factorise <math>x^2+2x+2</math>. The equation <math>x^2+2x+2=0</math> has the solutions
It now remains to factorise <math>x^2+2x+2</math>. The equation <math>x^2+2x+2=0</math> has the solutions
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{{Displayed math||<math>x=-1\pm \sqrt{\smash{(-1)^2 -2}\vphantom{i^2}} = -1 \pm \sqrt{-1} = -1\pm i</math>}}
+
{{Abgesetzte Formel||<math>x=-1\pm \sqrt{\smash{(-1)^2 -2}\vphantom{i^2}} = -1 \pm \sqrt{-1} = -1\pm i</math>}}
and therefore the polynomial has the following factorization into complex linear factors.
and therefore the polynomial has the following factorization into complex linear factors.
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{{Displayed math||<math>\begin{align*} x^3+x^2-2 = (x-1)(x^2+2x+2) &= (x-1)(x-(-1+i))(x-(-1-i))\\ &= (x-1)(x+1-i)(x+1+i)\,\mbox{.}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*} x^3+x^2-2 = (x-1)(x^2+2x+2) &= (x-1)(x-(-1+i))(x-(-1-i))\\ &= (x-1)(x+1-i)(x+1+i)\,\mbox{.}\end{align*}</math>}}
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Version vom 12:50, 10. Mär. 2009

       Theory          Exercises      

Contents:

  • Factor theorem.
  • Polynomial division
  • Fundamental theorem of algebra

Learning outcomes:

After this section, you will have learned to:

  • Perform polynomial division.
  • Understand the relationship between factors and zeros of polynomials.
  • Know that a polynomial equation of degree n has n roots (including multiplicity).
  • Know that real polynomial equations have complex conjugate roots.

Polynomials and equations

An expression of the form

\displaystyle a_nx^n+a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x+a_0

where \displaystyle n is an integer, is called a polynomial of degree \displaystyle n in an unknown variable \displaystyle x. The number \displaystyle a_1 is called the coefficient of \displaystyle x, \displaystyle a_2 the coefficient of \displaystyle x^2, etc. The constant \displaystyle a_0 is called the constant term.


Polynomials are essential for a large part of mathematics and have many properties that display great similarities with integers expressed using our "Arabic" (actually originally Indian) place value system. This means that we can often do calculations with polynomials in a similar way as with integers.


Example 1


Compare the following integer written using a base 10,

\displaystyle 1353= 1\times 10^3 + 3\times 10^2 + 5\times 10 + 3

with the polynomial in \displaystyle x

\displaystyle x^3 + 3x^2 + 5x + 3 = 1\times x^3 + 3\times x^2 + 5\times x + 3

and then the following divisions,

  • \displaystyle \quad\frac{1353}{11} = 123 \qquad as \displaystyle \ 1353= 123\times 11\,,
  • \displaystyle \quad\frac{x^3 + 3x^2 + 5x + 3}{x+1} = x^2+2x+3\qquad since \displaystyle \ x^3 + 3x^2 + 5x + 3= (x^2+2x+3)(x+1)\,.

If \displaystyle p(x) is a polynomial of degree \displaystyle n then \displaystyle p(x)=0 is called a polynomial equation of degree \displaystyle n. If \displaystyle x=a is a number such that \displaystyle p(a)=0 then \displaystyle x=a is called a root, or solution of the equation. One also says that \displaystyle x=a is a zero of \displaystyle p(x).

As the above example showed, polynomials can be divided very like integers. Polynomial division, like integer division, is usually not exact. If, for example, \displaystyle 37 is divided by \displaystyle 5, one gets

\displaystyle \frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}

The calculation can also be written as \displaystyle \ 37= 7\times 5+2\,. The number 7 is called the quotient and the number 2 the remainder. One says that dividing 37 by 5 gives the quotient 7 and the remainder 2.


Similarly, if \displaystyle p(x) and \displaystyle q(x) are polynomials, one can divide \displaystyle p(x) by \displaystyle q(x) and unambiguously determine polynomials \displaystyle k(x) and \displaystyle r(x) such that

\displaystyle \frac{p(x)}{q(x)} = k(x)+ \frac{r(x)}{q(x)}\,\mbox{,}

or \displaystyle \ p(x)= k(x)\, q(x)+r(x)\,. One here says that polynomial division has resulted in a quotient \displaystyle k(x) and remainder \displaystyle r(x).


Clearly, a division is exact if the remainder is zero. For polynomials this is expressed as follows: If \displaystyle r(x)=0 then \displaystyle p(x) is divisible by \displaystyle q(x), or, \displaystyle q(x) is a divisor of \displaystyle p(x). One writes

\displaystyle \frac{p(x)}{q(x)} = k(x)\,\mbox{,}

or \displaystyle \ p(x) = k(x)\, q(x)\,.


Polynomial division

If \displaystyle p(x) is a polynomial of higher degree than polynomial \displaystyle q(x) then one can divide \displaystyle p(x) by \displaystyle q(x). For example, this may be done by gradually subtracting appropriate multiples of \displaystyle q(x) from \displaystyle p(x) until a remaining numerator is of lower degree than the denominator \displaystyle q(x).


Example 2


Perform polynomial divisionen for \displaystyle \ \frac{x^3 + x^2 -x +4}{x+2}\,.


The first step is that we add and subtract an appropriate \displaystyle x^2-term in the numerator

\displaystyle \frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}

The reason why we do this is that the sub-expression \displaystyle x^3+2x^2 can be written as \displaystyle x^2(x+2) and cancellation with the denominator can be done,

\displaystyle \frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}

Then we add and subtract an appropriate \displaystyle x-term so that the leading \displaystyle x^2-term in the numerator can be cancelled,

\displaystyle \begin{align*} x^2+\frac{-x^2-2x+2x-x+4}{x+2} &= x^2+\frac{-x(x+2)+2x-x+4}{x+2}\\ &=x^2-x+\frac{x+4}{x+2}\,\mbox{.}\end{align*}

The last step is that we add and subtract a constant

\displaystyle x^2-x+\frac{x+4}{x+2}=x^2-x+\frac{x+2-2+4}{x+2} = x^2-x+1+\frac{2}{x+2}\,\mbox{.}

Thus ending up with

\displaystyle \frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}

The quotient is \displaystyle x^2 -x + 1 and the remainder is \displaystyle 2. Since the remainder is not zero, division is not exact, that is, \displaystyle q(x)= x+2 is not a divisor of \displaystyle p(x)=x^3 + x^2 -x +4.


The connection between factors and zeros

If \displaystyle q(x) is a divisor of \displaystyle p(x) then \displaystyle p(x)=k(x)\, q(x). We have thus factorised \displaystyle p(x) . One says that \displaystyle q(x) is a factor of \displaystyle p(x). Especially, if a polynomial of first degree \displaystyle (x-a) is a divisor of \displaystyle p(x) then \displaystyle (x-a) is a factor of \displaystyle p(x) , i.e.

\displaystyle p(x)= q(x)\, (x-a)\,\mbox{.}

Since \displaystyle \ p(a)=q(a)\, (a-a)= q(a)\times 0 = 0\ this means that \displaystyle x=a is a zero of \displaystyle p(x). This is exactly the content of the so-called factor theorem.

Factor theorem:

\displaystyle (x-a) is a divisor of a polynomial \displaystyle p(x) if and only if \displaystyle x=a is a zero of \displaystyle p(x).

Please note that the theorem applies in both directions, i.e. if we know that \displaystyle x=a is a zero of \displaystyle p(x) we automatically would know that \displaystyle p(x) is divisible by \displaystyle (x-a).


Example 3


The polynomial \displaystyle p(x) = x^2-6x+8 can be factorised as

\displaystyle x^2-6x+8 = (x-2)(x-4)

and has therefore zeros at \displaystyle x=2 and \displaystyle x=4 (and no others). It is precisely these that are obtained if one solves the equation \displaystyle \ x^2-6x+8 = 0\,.

Example 4


  1. Factorise the polynomial \displaystyle \ x^2-3x-10\,.

    By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation \displaystyle \ x^2-3x-10=0\ has the solutions
    \displaystyle x= \frac{3}{2} \pm \sqrt{\Bigl(\frac{3}{2}\Bigr)^2 - (-10)} = \frac{3}{2} \pm \frac{7}{2}\,\mbox{,}

    i.e. \displaystyle x=-2 and \displaystyle x=5. This means that \displaystyle \ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,.

  2. Factorise the polynomial \displaystyle \ x^2+6x+9\,.

    This polynomial has a repeated root
    \displaystyle x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3

    and thus \displaystyle \ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,.

  3. Factorise the polynomial \displaystyle \ x^2 -4x+5\,.

    In this case, the polynomial has two complex roots
    \displaystyle x= 2 \pm \sqrt{2^2 -5} = 2\pm \sqrt{-1} = 2\pm i

    and when factorised will be \displaystyle \ (x-(2-i))(x-(2+i))\,.

Example 5


Determine a cubic polynomial having zeros, \displaystyle 1, \displaystyle -1 and \displaystyle 3.

According to the factor theorem, the polynomial must have factors \displaystyle (x-1), \displaystyle (x+1) and \displaystyle (x-3). Multiplying these factors, we get a cubic polynomial

\displaystyle (x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}


Fundamental theorem of algebra

At the beginning of this chapter, we introduced the complex numbers to enable us to solve the quadratic equation \displaystyle x^2=-1 and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or whether we need to invent more types of numbers in order to solve other complicated polynomials. The answer to that question is that we need not; the complex numbers are enough. The German mathematician Carl Friedrich Gauss proved in the year 1799 the fundamental theorem of algebra which says the following:

Every polynomial of degree \displaystyle n\ge1 with complex coefficients has at least one zero which is a complex number.

As every zero according to the the factor theorem is matched by a factor, we can now also state the following theorem:

Every polynomial of degree \displaystyle n\ge1 has exactly \displaystyle n zeros if each zero is counted up to its multiplicity.

(By multiplicity is meant that a double zero is counted twice, a triple zero 3 times, etc.)


Note that these theorems only say that there exist complex roots of a polynomial, but not how to determine them. In general, there is no simple method to write a formula for the roots, and for polynomials of higher degree, we must use various devices to obtain a solution. If we restrict ourselves to polynomial with real coefficients, one of the devices that can help us is the knowledge that the complex roots of such polynomials always come in complex conjugate pairs.


Example 6


Show that the polynomial \displaystyle p(x)=x^4-4x^3+6x^2-4x+5 has zeros \displaystyle x=i and \displaystyle x = 2-i. Thus determine the other zeros.


We have

\displaystyle \begin{align*} p(i) &= i^4- 4i^3 +6i^2-4i+5 = 1+4i-6-4i+5=0\,\mbox{,}\\ p(2-i) &= (2-i)^4 -4(2-i)^3 + 6(2-i)^2 - 4(2-i) +5\,\mbox{.}\end{align*}

In order to calculate the last term, we need to determine

\displaystyle \begin{align*} (2-i)^2 &= 4-4i+i^2 = 3-4i\,\mbox{,}\\ (2-i)^3 &= (3-4i)(2-i) = 6-3i-8i+4i^2 = 2-11i\,\mbox{,}\\ (2-i)^4 &= (2-11i)(2-i) = 4-2i-22i+11i^2= -7-24i\,\mbox{.}\end{align*}

This gives that

\displaystyle \begin{align*} p(2-i) &= -7-24i-4(2-11i)+6(3-4i) -4(2-i) +5\\ &= -7-24i-8+44i+18-24i-8+4i+5=0\,\mbox{,}\end{align*}

which proves that \displaystyle i and \displaystyle 2-i are zeros of this polynomial.


Since the polynomial has real coefficients, we can immediately say that the other two zeros are the complex conjugates of the first two zeros, i.e. the other two roots are \displaystyle z=-i and \displaystyle z=2+i.

One consequence of the fundamental theorem of algebra (and the factor theorem) is that all polynomials can be factored into a product of complex linear factors. This also applies to polynomials with real coefficients, but for such polynomials it is possible to multiply together the pair of factors belonging to complex conjugate roots. In this case the factorisation will consist of linear and quadratic real factors.


Example 7


Show that \displaystyle x=1 is a zero of \displaystyle p(x)= x^3+x^2-2. Then first factorise \displaystyle p(x) into polynomials having real coefficients and then factorise \displaystyle p(x) completely into linear factors.


We have that \displaystyle \ p(1)= 1^3 + 1^2 -2 = 0\ which shows that \displaystyle x=1 is a zero of the polynomial. According to the factor theorem, this means that \displaystyle x-1 is a factor of \displaystyle p(x), i.e. \displaystyle p(x) is divisible by \displaystyle x-1. We therefore divide the polynomial with \displaystyle x-1 to get the remaining factors after \displaystyle x-1 is factored out of the polynomial

\displaystyle \begin{align*} \frac{x^3+x^2-2}{x-1} &= \frac{x^2(x-1)+2x^2-2}{x-1} = x^2 + \frac{2x^2-2}{x-1} = x^2 + \frac{2x(x-1) +2x -2}{x-1}\\[4pt] &= x^2 + 2x + \frac{2x-2}{x-1} = x^2 + 2x + \frac{2(x-1)}{x-1} = x^2 + 2x + 2\,\mbox{.}\end{align*}

So we have \displaystyle \ p(x)= (x-1)(x^2+2x+2)\, which is the first part of the problem.


It now remains to factorise \displaystyle x^2+2x+2. The equation \displaystyle x^2+2x+2=0 has the solutions

\displaystyle x=-1\pm \sqrt{\smash{(-1)^2 -2}\vphantom{i^2}} = -1 \pm \sqrt{-1} = -1\pm i

and therefore the polynomial has the following factorization into complex linear factors.

\displaystyle \begin{align*} x^3+x^2-2 = (x-1)(x^2+2x+2) &= (x-1)(x-(-1+i))(x-(-1-i))\\ &= (x-1)(x+1-i)(x+1+i)\,\mbox{.}\end{align*}
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