Find all rational solutions of x4−6x3+22x2−30x+13=0
x4−6x3+22x2−30x+13=0x^4−6x^3+22x^2−30x+13=0x4−6x3+22x2−30x+13=0
Let f(x) = x4−6x3+22x2−30x+13
f(1)=1−6+22−30+13=0f(1) = 1-6+22-30+13=0f(1)=1−6+22−30+13=0
So, (x−1)(x-1)(x−1) is a factor of f(x).f(x).f(x).
So, x4−6x3+22x2−30x+13=(x−1)x^4−6x^3+22x^2−30x+13 = (x-1)x4−6x3+22x2−30x+13=(x−1) (x3−5x2+17x−13x^3-5x^2+17x-13x3−5x2+17x−13 )
Let g(x)=x3−5x2+17x−13g(x)=x^3-5x^2+17x-13g(x)=x3−5x2+17x−13
g(1)=1−5+17−13=0g(1)=1-5+17-13=0g(1)=1−5+17−13=0
So, (x−1)(x-1)(x−1) is a factor of g(x).g(x).g(x).
Thus, f(x)=(x−1).g(x)=(x−1)(x−1)(x2−4x+13)f(x)=(x-1).g(x)=(x-1)(x-1)(x^2-4x+13)f(x)=(x−1).g(x)=(x−1)(x−1)(x2−4x+13)
Now, put (x2−4x+13)=0(x^2-4x+13)=0(x2−4x+13)=0
Using Quadratic formula:
x=4±16−522=4±6i2=2±3ix=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm6i}{2}=2\pm3ix=24±16−52=24±6i=2±3i
Hence, the solution are x=1,1,2+3i,2−3ix=1,1,2+3i,2-3ix=1,1,2+3i,2−3i and rational solutions are x=1,1x=1,1x=1,1
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