Find a polynomial function Px of least possible degree havin

Find a polynomial function P(x) of least possible degree, having real coefficients, with the given zeros. Write P(x) in expanded form. - 1 (with multiplicity 2), 4i (2 + i), 5 - 1, 1 + Squareroot 3, 1 - Squareroot 3 3, 1 (multiplicity 2); P(2) = - 5

Solution

Let a be an arbitrary real number. Then

a) P(x) = a(x+1)²( x-4i) = a( x²+2x+1) (x -4i) = a[ x³ +(2-4i)x²+ (1-8i)x -4i]

b) P(x) =a[(x-5){x-(2+i)(x-(2-i)}] ( complex roots occur in conjugate pairs) = (x-5)[ x² -(2+i)²] = a(x-5)[ x² -( 4+2i +i²)] =  a(x-5)[ x² - (3+2i )] (as i² = -1) = a[ x³ -5x²-(3+2i)x+ (15+10i)]

c) P(x)= a(x+1)[ {x - (1+3)}{( x- (1-3) }] = a(x+1)( x² -2x-2) = a( x³ -x² -4x -2).

d) P(x) = a (x-3)(x-1)² = a( x-3)( x² -2x +1) =a( x³ -5x² +7x -3). Now, since P(2) =-5, on substituting x = 2, we have -5 = a( 2³ -5*2²+7*2 -3) = a( 8-20+14 -3) = -a so that a = 5. Then P(x) = 5 (x³ -5x² +7x -3) = 5x³ -25x²+35x -15.