Chapter 6 Rational Exponents and Radical Functions - 6.5 Graph Square Root and Cube Root Functions - 6.5 Exercises - Mixed Review - Page 451: 49

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Given: $f(x)=x^4+x^3+2x^2+x-8$ The possible rational zeros are: $\pm 1, \pm 2, \pm 4, \pm 8$ The given equation becomes: $(x-1)(x^3+2x^2+4x+8)=0\\(x-1)[(x^3+2x^2)+(4x+8)]=0\\(x-1)[(x^2+4)(x+2)]=0\\(x-1)(x+2)(x-2i)(x+2i)=0$ The solutions are: $x=1\\x=-2\\x=2i\\x=-2i$