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Contents - Student subdomain for University of Bath

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56 CHAPTER 3. POLYNOMIAL EQUATIONS<br />

However, if b 2 − 4ac = 0, i.e. c = b 2 /4a then there is only one solution: x = −b<br />

In this case, the equation becomes ax 2 + bx + b2<br />

4a<br />

= 0, which can be re-written<br />

as a ( x +<br />

2a) b 2<br />

= 0, making it more obvious that there is a repeated root, and<br />

that the polynomial is not square-free (definition 32).<br />

Mathematicians dislike the sort <strong>of</strong> anomaly in “this equations has two solutions<br />

except when c = b 2 /4a”, especially as there are two roots as c tends to<br />

the value b 2 /4a. We there<strong>for</strong>e say that, in this special case, x = −b<br />

2a<br />

is a double<br />

root <strong>of</strong> the equation. This can be generalised, and made more <strong>for</strong>mal.<br />

2a .<br />

Definition 33 If, in the equation f = 0, f has a square-free decomposition<br />

f = ∏ n<br />

i=1 f i i, and x = α is a root <strong>of</strong> f i, we say that x = α is a root <strong>of</strong> f <strong>of</strong><br />

multiplicity i. When we say we are counting the roots <strong>of</strong> f with multiplicity,<br />

we mean that x = α should be counted i times.<br />

Proposition 17 The number <strong>of</strong> roots <strong>of</strong> a polynomial equation over the complex<br />

numbers, counted with multiplicity, is equal to the degree <strong>of</strong> the polynomial.<br />

Pro<strong>of</strong>. deg(f) = ∑ ideg(f i ), and each root <strong>of</strong> f i is to be counted i times as a<br />

root <strong>of</strong> f. That f i has i roots is the so-called Fundamental Theorem <strong>of</strong> Algebra.<br />

In this case, the two roots are given by the two possible signs <strong>of</strong> the square<br />

root, and √ 0 is assumed to have both positive and negative signs.<br />

3.1.2 Cubic Equations<br />

There is a <strong>for</strong>mula <strong>for</strong> the solutions <strong>of</strong> the cubic equation<br />

x 3 + ax 2 + bx + c, (3.6)<br />

albeit less well-known to schoolchildren:<br />

√<br />

1 3<br />

36 ba − 108 c − 8 a<br />

6<br />

3 + 12 √ 12 b 3 − 3 b 2 a 2 − 54 bac + 81 c 2 + 12 ca 3 −<br />

2b − 2 3<br />

√ a2<br />

36 ba − 108 c − 8 a3 + 12 √ 12 b 3 − 3 b 2 a 2 − 54 bac + 81 c 2 + 12 ca − 1 3 3 a.<br />

3<br />

We can simplify this by making a trans<strong>for</strong>mation 1 to equation (3.6): replacing<br />

x by x − a 3<br />

. This trans<strong>for</strong>ms it into an equation<br />

x 3 + bx + c (3.7)<br />

(where b and c have changed). This has solutions <strong>of</strong> the <strong>for</strong>m<br />

√<br />

1 3<br />

−108 c + 12 √ 12 b<br />

6<br />

3 + 81 c 2 −<br />

2b<br />

√ √<br />

−108 c + 12 12 b3 + 81 c . (3.8)<br />

2<br />

1 This is the simplest case <strong>of</strong> the Tschirnhaus trans<strong>for</strong>mation[vT83], which can always<br />

eliminate the x n−1 term in a polynomial <strong>of</strong> degree n.<br />

3

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