Let f(x)=x 6 +x 3 +1 Note: refer to second picture for "part (b)" a=-1 → a = 1.67a =0 >C+b =1> et1=1 > e=0.dte = 0dt1 =0in 72.foo is reducible the onlya (b-d) =0, b-d =0. > b =dabtetl=0.atb...

Let
f(x)=x⁶+x³+1

Note: refer to second picture for "part (b)"

Extracted text: Suppose that foo is reducible the only Given fo) =xf+x1 We hare to Show that chich is mposi ble. Themefone. foo # (x?+0x+ bx+1)(x&+dx'4ex+1, Case 1: foo is imeducible orer Let fo) = (4ax+1)(4bx4ex4chet1). X4 (a+b)x²+(1+ab te)xt(ae +b#d)x°+(e4adt) + (atd)x+1. Case I : Suppose hat fw= (x+1)(270x4 bx2te7 duc+1) atb =0 1+ab+e =0 +1. aetbtd =1, atd =0 both side cse get. Ctadt1 =0 , #3 ={0,23. Suppose that possi ble factors of foo are.. Compaining atlzŏ > a=-1 → a = 1. 67a =0 > C+b =1> et1=1 > e=0. dte = 0 dt1 =0 in 72. foo is reducible the only a (b-d) =0, b-d =0. > b =d abtetl=0. atb =0. hich is mpoesible, b= -a → b=-| > b=1. "Thene fone. fo) + farti) (x4 be?tex^4 dx+4) Suppose, for) = bêtoxt beti) (x?+ dh4ex+1). - d=0. d= -1 d=D1. Case 2 te)= (taxFbe+1) (x?+dx4 ex+1). or fe) = (x+1) (x?t+ bx?+ ex4 dx+1) or 7(24x41)= x4(a+d)x°+(e+od+b)x+ (ae+2+kd)x (ebtdlia)x "henefone [d=0], d=4] 3. d= 4 + atd =0 , actbd+2=1, etadtb=0 +(bte)k +1. ahich is Not possible. Cbtdta =0, Thenefione fe) + Cuti) (x4 on1+ bx4c4da+1. Hence fa) is îmeducible OVVER Z,


Extracted text: (1/6)æ° + (2/3)x³ + (4/3)x* + (7/6)æ³ + (5/3)x² + Use part (b) to prove that g(x) (2/3)x + (5/6) is irreducible over Q. ||