Find all other zeros of Px x3 9x2 36x 54 given that 3 3

Find all other zeros of P(x) = x^3 - 9x^2 + 36x - 54, given that 3 - 3i is a zero. (If there is mote than one zero, separate them with commas.)

Solution

Let P (x) = [ x - (3 -3i)] (ax² + bx +c) where a, b, c are constants. Then ax³ + [ b - (3-3i)] x² + [ c - (3-3i)b]x - (3- 3i)c = x³ -9x² + 36 x - 54, so that a = 1, b - (3-3i) = -9, c -(3-3i)b = 36 and (3- 3i)c = 54. Thus, a = 1, b = - 6 - 3i and c = 9 + 9i. Thus, the other zerof of P( x ) are the xeros of x² - (6 +3i)x + (9 +9i) = 0. On using the quadratic formula, we get x = [ -( -6 - 3i) ± { ( -6-3i)² -4*1(9+ 9i)}]/ 2*1 = [ (6 + 3i) ± { (36 + 36i + 9i²) -36 -36i} ]/2 = [(6 + 3i)± -9]/2 = [ (6 + 3i) ± 3i]/2 Thus, x = 3 and x = 3 + 3i are the other zeros of P(x).

Note:

Since complex roots always occur in conjugate pairs,3 +3i is also a zero.Then P (x)= [(x -( 3 -3i)] [x -(3+ 3i]( x -p).= x³ -9x² + 36 x - 54. We can solve this equation and get p = 3. This is a shorter method.