Chapter 5 Polynomials and Polynomial Functions - 5.7 Apply the Fundamental Theorem of Algebra - 5.7 Exercises - Problem Solving - Page 386: 65

$x\approx 4.25, 12.3$

Volume of the pyramid is given by $V_1=\dfrac{1}{3}(2x)(2x)(x)=\dfrac{4x^3}{3}$ Volume of the base $V_2=(2x+6)(2x+6)x=4x^3+24x^2+36x$ Now, total volume is $V=V_1+V_2$ $4x^3+24x^2+36x+\dfrac{4x^3}{3}=1000$ or, $4x^3+18x^2+27x-750=0$ or, $(x-4.25)(4x^2-35x+175)=0$ or, $(x-4.25)(x+3.55)(4x-49.2)=0$ Thus, $x\approx 4.25, 12.3$