IDENTITIES INVOLVING THE PRODUCT OF THREE BINOMIALS

Question 1 :

Using algebraic identity, find the coefficients of x² , x and constant term without actual expansion.

(i) (x + 5)(x + 6)(x + 7)

Solution :

(x + 5)(x + 6)(x + 7)

x = x, a = 5, b = 6 and c = 7

= x³ + (a + b + c) x² + (ab + bc + ca)x + abc

a + b + c  =  5 + 6 + 7  =  18

ab + bc + ca  =  5(6) + 6(7) + 7(5)

=  30 + 42 + 35

=  107

abc  =  5(6)(7)  =  210

= x³ + 18x² + 107x + 210

Coefficient of x²  =  18

Coefficient of x  =  107

Constant term  =  210

(ii) (2x + 3)(2x −5)(2x −6)

Solution :

 (2x + 3)(2x −5)(2x −6)

x = 2x, a = 3, b = -5 and c = -6

= x³ + (a + b + c) x² + (ab + bc + ca)x + abc

a + b + c  =  3 + (-5) + (-6)  =  -8

ab + bc + ca  =  3(-5) + (-5)(-6) + (-6)(3)

=  -15 + 30 - 18

=  -3

abc  =  3(-5)(-6)  =  90

= x³ - 8x² - 3x + 90

= (2x)³ - 8(2x)² - 3(2x) + 90

=  8x³ - 32x² - 6x + 90

Coefficient of x²  =  -32

Coefficient of x  =  -6

Constant term  =  90

Question 2 :

If (x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70 , find the value of

(i)  a + b + c

(ii)  (1/a) + (1/b) + (1/c)

(iii)  a² + b² + c²

(iv)  (a/bc) + (b/ac) + (c/ab)

Solution :

(x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70

x³ + (a+b+c)x² + (ab+bc+ca)x + abc  =  x³ + 14x² + 59x + 70

By equating the coefficient of x², we get

a + b + c  =  14

By equating the coefficient of x, we get

ab + bc + ca  =  59

By equating constants, we get

abc  =  70

(ii)  (1/a) + (1/b) + (1/c)

By taking L.C.M, we get

(1/a) + (1/b) + (1/c)  =  (bc + ac + ab)/abc

=  59/70

(iii)  a² + b² + c²

a² + b² + c²  =  (a+b+c)² - 2(ab + bc +ca)

  =  14² - 2(59)

  =   196 - 118

  =  78

(iv)  (a/bc) + (b/ac) + (c/ab)

By taking L.C.M, we get

  =  (a² + b² + c²)/abc

  =  78/70

  =  39/35

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