5.1. POLYNOMIALS AND ALGEBRAIC EQUATIONS 157 To find the coefficients b 0 , , b n of this expansion, one first divides f(x)byx – c with remainder. The remainder is b 0 , andthe quotient issome polynomial g 0 (x). Then one divides g 0 (x)byx – c with remainder. The remainder is b 1 , and the quotient is some polynomial g 1 (x). Then one divides g 1 (x)byx – c, obtaining the coefficient b 2 as the remainder, etc. It is convenient to perform the computations by Horner’s scheme (see Paragraph 5.1.1-3). Example 4. Expand the polynomial f (x)=x 4 – 5x 3 – 3x 2 + 9 in powers of the difference x – 3 (c = 3). We write out Horner’s scheme, where the first row contains the coefficients of the polynomial f(x), the second row contains the coefficients of the quotient g 0 (x) and the remainder b 0 obtained when dividing f(x) by x – 3, the third row contains the coefficients of the quotient g 1 (x) and the remainder b 1 obtained when dividing g 0 (x)byx – 3,etc.: 1 –5 –30 9 3 1 –2 –9 –27 –72 11–6 –45 14 6 17 1 Thus the expansion of f(x)inpowersofx – 3 has the form f(x)=(x – 3) 4 + 7(x – 3) 3 + 6(x – 3) 2 – 45(x – 3)–72. The coefficients in the expansion of a polynomial f (x) in powers of the difference x – c are related to the values of the polynomial and its derivatives at x = c by the formulas b 0 = f (c), b 1 = f  x (c) 1! , b 2 = f  xx (c) 2! , , b n = f (n) x (c) n! , where the derivative of a polynomial f (x)=a n x n + a n–1 x n–1 + ···+ a 1 x + x 0 with real or complex coefficients a 0 , , a n is the polynomial f  x (x)=na n x n–1 +(n–1)a n–1 x n–2 +···+a 1 , f  xx (x)=[f  x (x)]  x , etc. Thus Horner’s scheme permits one to find the values of the derivatives of the polynomial f (x)atx = c. Example 5. In Example 4, the values of the derivatives of the polynomial f(x)atx = 3 are f(3)=–72, f  (3)=–45 × 1!=–45, f  (3)=6×2!=12, f  (3)=7×3!=42, f IV (3)=1×4!=24. The expansion of a polynomial in powers of x–c can be used to compute the partial frac- tion decomposition of a rational function whose denominator is a power of a linear binomial. Example 6. Find the partial fraction decomposition of the rational function Φ(x)= x 2 + x + 1 (x – 2) 4 . First, we expand the polynomial f(x)=x 2 + x + 1 in powers of the binomial x –(–2)=x + 2: 111 –2 1 –13 1 –3 1 Thus f(x)=(x + 2) 2 – 3(x + 2)+3. As a result, we obtain Φ(x)= f(x) (x + 2) 4 = 1 (x + 2) 2 – 3 (x + 2) 3 + 3 (x + 2) 4 . 5.1.2. Linear and Quadratic Equations 5.1.2-1. Linear equations. The linear equation ax + b = 0 (a ≠ 0) has the solution x =– b a . 158 ALGEBRA 5.1.2-2. Quadratic equations. The quadratic equation ax 2 + bx + c = 0 (a ≠ 0)(5.1.2.1) has the roots x 1,2 = –b √ b 2 – 4ac 2a . The existence of real or complex roots is determined by the sign of the discriminant D = b 2 – 4ac: Case D > 0. There are two distinct real roots. Case D < 0. There are two distinct complex conjugate roots. Case D = 0. There are two equal real roots. V I ` ETE THEOREM. The roots of a quadratic equation (5.1.2.1) satisfy the following relations: x 1 + x 2 =– b a , x 1 x 2 = c a . 5.1.3. Cubic Equations 5.1.3-1. Incomplete cubic equation. 1 ◦ . Cardano’s solution. The roots of the incomplete cubic equation y 3 + py + q = 0 (5.1.3.1) have the form y 1 = A + B, y 2,3 =– 1 2 (A + B) i √ 3 2 (A – B), where A =  – q 2 + √ D  1/3 , B =  – q 2 – √ D  1/3 , D =  p 3  3 +  q 2  2 , i 2 =–1, and A, B are arbitrary values of the cubic roots such that AB =– 1 3 p. The number of real roots of a cubic equation depends on the sign of the discriminant D: Case D > 0. There is one real and two complex conjugate roots. Case D < 0. There are three real roots. Case D = 0. There is one real root and another real root of double multiplicity (this case is realized for p = q = 0). 2 ◦ . Trigonometric solution. If an incomplete cubic equation (5.1.3.1) has real coefficients p and q, then its solutions can be found with the help of the trigonometric formulas given below. (a)Letp < 0 and D < 0.Then y 1 = 2  – p 3 cos α 3 , y 2,3 =–2  – p 3 cos  α 3 π 3  , where the values of the trigonometric functions are calculated from the relation cos α =– q 2  –(p/3) 3 . 5.1. POLYNOMIALS AND ALGEBRAIC EQUATIONS 159 (b)Letp > 0 and D ≥ 0.Then y 1 = 2  p 3 cot(2α), y 2,3 =  p 3  cot(2α) i √ 3 sin(2α)  , where the values of the trigonometric functions are calculated from the relations tan α =  tan β 2  1/3 ,tanβ = 2 q  p 3  3/2 , |α| ≤ π 4 , |β| ≤ π 2 . (c)Letp < 0 and D ≥ 0.Then y 1 =–2  – p 3 1 sin(2α) , y 2,3 =  – p 3  1 sin(2α) i √ 3 cot(2α)  , where the values of the trigonometric functions are calculated from the relations tan α =  tan β 2  1/3 ,sinβ = 2 q  – p 3  3/2 , |α| ≤ π 4 , |β| ≤ π 2 . In the above three cases, the real value of the cubic root should be taken. 5.1.3-2. Complete cubic equation. The roots of a complete cubic equation ax 3 + bx 2 + cx + d = 0 (a ≠ 0)(5.1.3.2) are calculated by the formulas x k = y k – b 3a , k = 1, 2, 3, where y k are the roots of the incomplete cubic equation (5.1.3.1) with the coefficients p =– 1 3  b a  2 + c a , q = 2 27  b a  3 – bc 3a 2 + d a . V I ` ETE THEOREM. The roots of a complete cubic equation (5.1.3.2) satisfy the following relations: x 1 + x 2 + x 3 =– b a , x 1 x 2 + x 1 x 3 + x 2 x 3 = c a , x 1 x 2 x 3 =– d a . 5.1.4. Fourth-Degree Equation 5.1.4-1. Special cases of fourth-degree equations. 1 ◦ .Thebiquadratic equation ax 4 + bx 2 + c = 0 can be reduced to a quadratic equation (5.1.2.1) by the substitution ξ = x 2 . Therefore, the roots of the biquadratic equations are given by x 1,2 =  –b + √ b 2 – 4ac 2a , x 3,4 =  –b – √ b 2 – 4ac 2a . 160 ALGEBRA 2 ◦ .Thereciprocal (algebraic) equation ax 4 + bx 3 + cx 2 + bx + a = 0 can be reduced to a quadratic equation by the substitution y = x + 1 x . The resulting quadratic equation has the form ay 2 + by + c – 2a = 0. 3 ◦ .Themodified reciprocal equation ax 4 + bx 3 + cx 2 – bx + a = 0 can be reduced to a quadratic equation by the substitution y = x – 1 x . The resulting quadratic equation has the form ay 2 + by + 2a + c = 0. 4 ◦ .Thegeneralized reciprocal equation ab 2 x 4 + bx 3 + cx 2 + dx + ad 2 = 0 can be reduced to a quadratic equation by the substitution y = bx + d x . The resulting quadratic equation has the form ay 2 + y + c – 2abd = 0. 5.1.4-2. General fourth-degree equation. 1 ◦ . Reduction of a general equation of fourth-degree to an incomplete equation.The general equation of fourth-degree ax 4 + bx 3 + cx 2 + dx + e = 0 (a ≠ 0) can be reduced to an incomplete equation of the form y 4 + py 2 + qy + r = 0 (5.1.4.1) by the substitution x = y – b 4a . 2 ◦ . Descartes–Euler solution. The roots of the incomplete equation (5.1.4.1) are given by the formulas y 1 = 1 2  √ z 1 + √ z 2 + √ z 3  , y 2 = 1 2  √ z 1 – √ z 2 – √ z 3  , y 3 = 1 2  – √ z 1 + √ z 2 – √ z 3  , y 4 = 1 2  – √ z 1 – √ z 2 + √ z 3  , (5.1.4.2) where z 1 , z 2 , z 3 are the roots of the cubic equation (cubic resolvent of equation (5.1.4.1)) z 3 + 2pz 2 +(p 2 – 4r)z – q 2 = 0.(5.1.4.3) The signs of the roots in (5.1.4.2) are chosen from the condition √ z 1 √ z 2 √ z 3 =–q. The roots of the fourth-degree equation (5.1.4.1) are determined by the roots of the cubic resolvent (5.1.4.3); see Table 5.1. 5.1. POLYNOMIALS AND ALGEBRAIC EQUATIONS 161 3 ◦ . Ferrari solution.Letz 0 be any of the roots of the auxiliary cubic equation (5.1.4.3). Then the four roots of the incomplete equation (5.1.4.1) are found by solving the following two quadratic equations: y 2 – √ z 0 y + p + z 0 2 + q 2 √ z 0 = 0, y 2 + √ z 0 y + p + z 0 2 – q 2 √ z 0 = 0. TABLE 5.1 Relations between the roots of an incomplete equation of fourth-degree and the roots of its cubic resolvent Cubic resolvent (5.1.4.3) Fourth-degree equation (5.1.4.1) All roots are real and positive* Four real roots All roots are real: one is positive and two are negative* Two pairs of complex conjugate roots One real root and two complex conjugate roots Two real roots and two complex conjugate roots 5.1.5. Algebraic Equations of Arbitrary Degree and Their Properties 5.1.5-1. Simplest equations of degree n and their solutions. 1 ◦ .Thebinomial algebraic equation x n – a = 0 (a ≠ 0) has the solutions x k+1 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ a 1/n  cos 2kπ n + i sin 2kπ n  for a > 0, |a| 1/n  cos (2k + 1)π n + i sin (2k + 1)π n  for a < 0, where k = 0, 1, , n – 1 and i 2 =–1. 2 ◦ . Equations of the form x 2n + ax n + b = 0, x 3n + ax 2n + bx n + c = 0, x 4n + ax 3n + bx 2n + cx n + d = 0 are reduced by the substitution y = x n to a quadratic, cubic, and fourth-degree equation, respectively, whose solution can be expressed by radicals. Remark. In the above equations, n can be noninteger. 3 ◦ .Thereciprocal (algebraic) equation a 0 x 2n + a 1 x 2n–1 + a 2 x 2n–2 + ···+ a 2 x 2 + a 1 x + a 0 = 0 (a 0 ≠ 0) can be reduced to an equation of degree n by the substitution y = x + 1 x . *BytheVi ` ete theorem, the product of the roots z 1 , z 2 , z 3 is equal to q 2 ≥ 0. 162 ALGEBRA Example 1. The equation ax 6 + bx 5 + cx 4 + dx 3 + cx 2 + bx + a = 0 can be reduced to the cubic equation ay 3 + by 2 +(c – 3a)y + d – 2b = 0 by the substitution y = x + 1/x. 5.1.5-2. Equations of general form and their properties. An algebraic equation of degree n has the form a n x n + a n–1 x n–1 + ···+ a 1 x + a 0 = 0 (a n ≠ 0), (5.1.5.1) where a k are real or complex coefficients. Denote the polynomial of degree n on the right-hand side in equation (5.1.5.1) by P n (x) ≡ a n x n + a n–1 x n–1 + ···+ a 1 x + a 0 (a n ≠ 0). (5.1.5.2) Avaluex = x 1 such that P n (x 1 )=0 is called a root of equation (5.1.5.1) (and also a root of the polynomial P n (x)). A value x = x 1 is called a root of multiplicity m if P n (x)=(x–x 1 ) m Q n–m (x), where m is an integer (1 ≤ m ≤ n), and Q n–m (x) is a polynomial of degree n – m such that Q n–m (x 1 ) ≠ 0. T HEOREM 1(FUNDAMENTAL THEOREM OF ALGEBRA). Any algebraic equation of degree n has exactly n roots (real or complex), each root counted according to its multiplicity. Thus, the left-hand side of equation (5.1.5.1) with roots x 1 , x 2 , , x s of the respective multiplicities k 1 , k 2 , , k s (k 1 + k 2 + ···+ k s = n) can be factorized as follows: P n (x)=a n (x – x 1 ) k 1 (x – x 2 ) k 2 (x – x s ) k s . T HEOREM 2. Any algebraic equation of an odd degree with real coefficients has at least one real root. THEOREM 3. Suppose that equation (5.1.5.1) with real coefficients has a complex root x 1 = α + iβ . Then this equation has the complex conjugate root x 2 = α – iβ , and the roots x 1 , x 2 have the same multiplicity. THEOREM 4. Any rational root of equation (5.1.5.1) with integer coefficients a k is an irreducible fraction of the form p/q ,where p is a divisor of a 0 and q is a divisor of a n .If a n = 1 , then all rational roots of equation (5.1.5.1) (if they exist) are integer divisors of the free term. THEOREM 5(ABEL–RUFFINI THEOREM). Any equation (5.1.5.1) of degree n ≤ 4 is solvable by radicals, i.e., its roots can be expressed via its coefficients by the operations of addition, subtraction, multiplication, division, and taking roots (see Subsections 5.1.2– 5.1.4). In general, equation (5.1.5.1) of degree n > 4 cannot be solved by radicals. 5.1. POLYNOMIALS AND ALGEBRAIC EQUATIONS 163 5.1.5-3. Relations between roots and coefficients. Discriminant of an equation. VI ` ETE THEOREM. The roots of equation (5.1.5.1) (counted according to their multiplicity) and its coefficients satisfy the following relations: (–1) k a n–k a n = S k (k = 1, 2, , n), where S k are elementary symmetric functions of x 1 , x 2 , , x n : S 1 = n  i=1 x i , S 2 = n  1≤i<j x i x j , S 3 = n  1≤i<j<k x i x j x k , , S n = x 1 x 2 x n . Note also the following relations: (n – k)a n–k + k  j=1 a n–(k–j) s j = 0 (k = 1, 2, , n) with symmetric functions s j = n  i=1 x j i . The discriminant D of an algebraic equation is the product of a 2n–2 n and the squared Vandermonde determinant Δ(x 1 , x 2 , , x n )ofitsroots: D = a 2n–2 n [Δ(x 1 , x 2 , , x n )] 2 = a 2n–2 n  1≤j<i≤n (x i – x j ) 2 . The discriminant D is a symmetric function of the roots x 1 , x 2 , , x n , and is equal to zero if and only if the polynomial P n (x) has at least one multiple root. 5.1.5-4. Bounds for the roots of algebraic equations with real coefficients. 1 ◦ . All roots of equation (5.1.5.1) in absolute value do not exceed N = 1 + A |a n | ,(5.1.5.3) where A is the largest of |a 0 |, |a 1 |, , |a n–1 |. The last result admits the following generalization: all roots of equation (5.1.5.1) in absolute value do not exceed N 1 = ρ + A 1 |a n | ,(5.1.5.4) where ρ > 0 is arbitrary and A 1 is the largest of |a n–1 |, |a n–2 | ρ , |a n–3 | ρ 2 , , |a 0 | ρ n–1 . For ρ = 1, formula (5.1.5.4) turns into (5.1.5.3). Remark. Formulas (5.1.5.3) and (5.1.5.4) can also be used for equations with complex coefficients. Example 2. Consider the following equation of degree 4: P 4 (x)=9x 4 – 9x 2 – 36x + 1. Formula (5.1.5.3) for n = 4, |a n | = 9, A = 36 yields a fairly rough estimate N = 5, i.e., the roots of the equation belong to the interval [–5, 5]. Formula (5.1.5.4) for ρ = 2, n = 4, |a n | = 9, A 1 = 9 yields a better estimate for the bounds of the roots of this polynomial, N 1 = 3. . expansion of a polynomial in powers of x–c can be used to compute the partial frac- tion decomposition of a rational function whose denominator is a power of a linear binomial. Example 6. Find the partial. 6 17 1 Thus the expansion of f(x)inpowersofx – 3 has the form f(x)=(x – 3) 4 + 7(x – 3) 3 + 6(x – 3) 2 – 45(x – 3)–72. The coefficients in the expansion of a polynomial f (x) in powers of the difference. =  p 3  3 +  q 2  2 , i 2 =–1, and A, B are arbitrary values of the cubic roots such that AB =– 1 3 p. The number of real roots of a cubic equation depends on the sign of the discriminant D: Case