Chapter 10 Counting Methods and Probability - 10.6 Construct and Interpret Binomial Distributions - 10.6 Exercises - Mixed Review - Page 730: 65

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$$f(x)=2x^4+5x^3+29x^2+80x-48=0\\2x^4+6x^3-x^3-3x^2+32x^2+96x-16x-48=0\\2x^3(x+3)-x^2(x+3)+32x(x+3)-16(x+3)=0\\(x+3)(2x^3-x^2+32x-16)=0\\(x+3)(2x-1)(x^2+16)=0$$ The solutions are $x+3=0\rightarrow x=-3\\2x-1=0 \rightarrow x=\frac{1}{2}$