Q13/046) 16x^4−mx^3+(2m+17)x^2−mx+16=0 has four distinct roots forming a geometric progression. Find all values of m.

Let the roots be a/r^3, a/r, ar and ar^3 (the common ratio = r^2)

product of roots = a^4 = 16/16 = 1

so the polynomial is q( x +1/r^3)(x+1/r)(x)(x+r)(x+ r^3)

or n( r^3 x + 1)(r x + 1) ( x + r) (x+r^3)

as the coefficients are symmetric we have n =1

now take the product to get r^4 = 16

r = 2 or -2(only possible values)

by putting 2 and then -2 we can get the result

r = 2 => (8x+1)(2x+1)(x+2)(x+8)

= ( 16x^2 + 10 x + 1)(x^2 + 10 x + 16)

= 16x^4 + 170 x^3 + 357x^2 + 170 x + 16

It does not satisfy

r = -2 => (8x-1)(2x-1)(x-2)(x-8)

= ( 16x^2 - 10 x + 1)(x^2 - 10 x + 16)

= 16x^4 - 170 x^3 + 357x^2 - 170 x + 16

m = 170 satisfies the original equation and hence m = 170 and solutions are 1/8. 1/2, 2, 8