Chapter 6 Rational Exponents and Radical Functions - 6.5 Graph Square Root and Cube Root Functions - 6.5 Exercises - Mixed Review - Page 451: 50

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Given: $f(x)=x^4+6x^3+14x^2+54x+45$ The possible rational zeros are: $\pm 1, \pm 3, \pm 5, \pm 9,\pm, \pm 45$ We found a zero: $x=-1$. The given equation becomes: $(x+1)(x^3+5x^2+9x+45)=0\\(x+1)[(x^3+5x^2)+(9x+45)]=0\\(x+1)(x^2+9)(x+5)=0\\(x+1)(x+5)(x-3i)(x+3i)=0$ The solutions are: $x=-1\\x=-5\\x=3i\\x=-3i$