Find all roots of polynomial using Bairstow method
`f(x)=9x^4+12x^3+13x^2+12x+4` and `r=-1, s=-1`

Solution:
`9x^4+12x^3+13x^2+12x+4=0`

Let the initial approximation be `r=-1` and `s=-1`

Here `a_4=9`, `a_3=12`, `a_2=13`, `a_1=12`, `a_0=4`

Iteration=1
`b_4=a_4=9`

`b_3=a_3+rb_4=12-1 * 9=3`

`b_2=a_2+rb_3+sb_4=13 -1 * 3 -1 * 9=1`

`b_1=a_1+rb_2+sb_3=12 -1 * 1 -1 * 3=8`

`b_0=a_0+rb_1+sb_2=4 -1 * 8 -1 * 1=-5`


`c_4=b_4=9`

`c_3=b_3+rc_4=3-1 * 9=-6`

`c_2=b_2+rc_3+sc_4=1 -1 * (-6) -1 * 9=-2`

`c_1=b_1+rc_2+sc_3=8 -1 * (-2) -1 * (-6)=16`


The simultaneous equations for `Deltar` and `Deltas` are

`c_2Deltar+c_3Deltas=-b_1` and `c_1Deltar+c_2Deltas=-b_0`

Substitute values of `c_1,c_2,c_3` and `b_0,b_1`

`-2Deltar-6Deltas=-8` and `16Deltar-2Deltas=5`

Solving equations using Cramer's rule method
`D=c_2*c_2-c_1*c_3=(-2)*(-2)-16*(-6)=100`

`D_1=b_0*c_3-b_1*c_2=(-5)*(-6)-8*(-2)=46`

`D_2=b_1*c_1-b_0*c_2=8*16-(-5)*(-2)=118`

`Deltar=D_1/D=(46)/(100)=0.46`

and `Deltas=D_2/D=(118)/(100)=1.18`

The new `r` and `s` are

`r=r+Deltar=-1+0.46=-0.54`

and `s=s+Deltas=-1+1.18=0.18`

The approximate error in r and s
`|epsilon_(a,r)|=|(Deltar)/(r)|xx100%=|(0.46)/(-0.54)|xx100%=85.1852`

`|epsilon_(a,s)|=|(Deltas)/(s)|xx100%=|(1.18)/(0.18)|xx100%=655.5556`

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Iteration=2
`b_4=a_4=9`

`b_3=a_3+rb_4=12-0.54 * 9=7.14`

`b_2=a_2+rb_3+sb_4=13 -0.54 * 7.14 +0.18 * 9=10.7644`

`b_1=a_1+rb_2+sb_3=12 -0.54 * 10.7644 +0.18 * 7.14=7.4724`

`b_0=a_0+rb_1+sb_2=4 -0.54 * 7.4724 +0.18 * 10.7644=1.9025`


`c_4=b_4=9`

`c_3=b_3+rc_4=7.14-0.54 * 9=2.28`

`c_2=b_2+rc_3+sc_4=10.7644 -0.54 * 2.28 +0.18 * 9=11.1532`

`c_1=b_1+rc_2+sc_3=7.4724 -0.54 * 11.1532 +0.18 * 2.28=1.8601`


The simultaneous equations for `Deltar` and `Deltas` are

`c_2Deltar+c_3Deltas=-b_1` and `c_1Deltar+c_2Deltas=-b_0`

Substitute values of `c_1,c_2,c_3` and `b_0,b_1`

`11.1532Deltar+2.28Deltas=-7.4724` and `1.8601Deltar+11.1532Deltas=-1.9025`

Solving equations using Cramer's rule method
`D=c_2*c_2-c_1*c_3=11.1532*11.1532-1.8601*2.28=120.1529`

`D_1=b_0*c_3-b_1*c_2=1.9025*2.28-7.4724*11.1532=-79.0038`

`D_2=b_1*c_1-b_0*c_2=7.4724*1.8601-1.9025*11.1532=-7.3193`

`Deltar=D_1/D=(-79.0038)/(120.1529)=-0.6575`

and `Deltas=D_2/D=(-7.3193)/(120.1529)=-0.0609`

The new `r` and `s` are

`r=r+Deltar=-0.54-0.6575=-1.1975`

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