The HCF and LCM of two quadratic expressions are respectively (x – 5) and x³ – 19x – 30, then find both the expressions.

HCF = (x – 5)

LCM = x³ – 19x – 30

= x³ – 19x – 38 + 8

= x³ + 8 – 19x – 38

= x³ + 2³ – 19(x + 2)

x³ + 2³ = (x + 2) (x² – 2x + 4) [Using a³ + b³ = (a + b) (a² – ab + b²)]

= (x + 2) (x² – 2x + 4) – 19(x + 2)

= (x + 2) (x² – 2x + 4 – 19)

= (x + 2) (x² – 2x – 15)

= (x + 2) (x² – 5x + 3x – 15)

= (x + 2) (x(x– 5) + 3(x – 5))

= (x + 2) (x – 5) (x + 3)

Since HCF = x – 5 it will belong to both polynomials.

So u(x) = (x – 5) (x + 3)

= x² – 2x - 15

So v(x) = (x – 5) (x + 2)

= x² – 3x - 10