Question

Use the given complex zero to find the remaining three zeros for the polynomial. Write the three values as a comma separated f(x)=x^-x3-8x2-44-48 has 21 as one of its zeros.

Answer 1

f(x) x x _x3_8x²_ux-48 given ( 21) is zero of jix) =) fc2i) root As a) "x38x² 2 we Now (21) is of fial (x - 2.1) is facter of x-x38x²4x-18 divided 82?_42-18 (x-2) 29+ (26-1)X?-(24+1272 - 240 (x-26) Jx4 -x38x3_48.48 x"_zix² (23-1) x3-82? (21-1) x3 +(4+2i) x² -(24+12)x3-47 - (2x+12)x+(-4+243)2 -24 ix-48 -241x-48 CS Scanned with CamScanner

2 3? - 8 x2-46-48 = (x-2)(**+ (21-1)* - (24+12)x-24 1] Now find root of x3+(21-1)x²_ (2i+12)x - 24 Jos = -3 thi's is -27+( 21-1) 9 (21+12) (-3) -241 -27 7181-9 + bit 36-zunzo O sate's fy * +126-1)x?_(2!+12)& x z-3 - 24 i CT is (36+ +3) Jecter of + c26-12x²_125+127x-2214 20 we Now divide +x+37 xx3+ (21-1) x² - (25+12)x -24; by (2+3). cs Scanned with CamScanner

x2 +(21-4) -si (x+3) 6737 (zuid)x+- (2x+12)2-zuk *3. + 3x² (23-4726?- (20+12) x (24-u)X? + (61-12) -gia-zui -fixa au i 2) x3+ (2x1)x² - (2i+12)x-zui (x+3)(c + (24-u) - 8i). in uring x"_x3_8x²_40-48 (x-26) (26+3) [x+12 0-4)-80] CS Scanned with CamScanner

3 Now we sind af. root 70? + (2004) x - 8ua Jos . x = 4 16+(21-44-dizo 16+8; -16- fi'z ó 0 jectes of x ²7 (21-47-81 3) (x-4) is divided now x²7 (28_47x - 8i by (21-4) we Ditzi (x-4) 12+123-4)-84 х-— Чx zix-81 > 2ix-81 = ) 2x²+(21-4)x - 8 = (x-4)(x+21) CS Scanned with CamScanner

in uring x3_800²_4 8 c²_40-48 (X-26) (3+3)(36-4)(x+21) root of f(x)=x"_x3_8x²_42-48 Hence is 20 = zi, -3, 4, -21 Hence the result. CS Scanned with CamScanner