Question

3. Any polynomial with real coefficients of degree k can be factored com- pletely into first-degree binomials which may include complex numbers. That is, for any real ao, Q1, ..., āk ao + a1x + a22² + ... + axxk = C(x – 21)(x – z2....(x – zk) for some real C and 21, 22, ... Zk possibly real or complex. Therefore, up to multiplicity, every polynomial of degree k has exactly k-many roots, includ- ing complex roots. Find all roots of the following polynomials: a. f(x) = x2 - 6x + 58 b. g(x) = x2 – 3x2 + 12x – 10 4. What do you notice about the complex roots of polynomials with real coefficients?

Answer 1

classmate Date Page (3) (a) Let Twe have to Sol2 f(a)=x² find the 6x+58 roots of from I function find the roots of the So consider, f(x)=0 7 - 6x + 58 = 0 T x = -(-) + c^ - 4( )(S 8 ) - 2012 The roots of ax² + bx to zo is given x = -b I 16²4ac - 2a - 6 £/36-232 - 6+ ]-196 = 6 7 14 i ( vizi & J196 2 = 2(3+77) 2 2 =14 स the Thus are roots of the polynomial F (a) x, = 3 +7 Ý 22 = 3-71 fra)=x²-3x² + 12x-10 (6) let So, consider f(2)=0=x²-3x²+12x-10 Clearly, (f(1) = a=1 is a rest of f(1) = 0 3-360327120.-10 = 1-3 +12-10=0) 1/3

classmate Date Page » >> (x-1) is a factor of the polynomial fox) = x²-3x² + 12x - 10 (-1) 7-3x+128-10 q? - 2 x tlo 7 -1 127-10 | +63 ? O -2742 +12% -10 7202t 24 10-10 #197 Flo O :.*?-34?+ 12% -10 = (2-1) (322+10) = (2+)(02-2x + 10) EO 3 2=1 of 12-24 tio o a=1 or a=-(-2) I 54-4(1)(10) 2. - 2 + 14-40 2 = 2 + 1 -36 2 e 2 + 6 2 √36=6 &&(1 + i) 2 = 1 tsi si 2/3

classmate Date Page Thus the roots of foo are &,= 1, da = 1t3i, 3 = polynoma the complex (4) We noticed that the in (a) , & (b) above have scots but occured in pairs ie. the complen roots are conjugate of each others ic if atiß is a root - d-iß is also the scot . a a | Hence, we can say that for any polynomial over R (ie with real coefficients if atiß is any roof then d-ips must be the another root. where a-iß is called the complex conjugate of a tiß. 3/3