Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 391: 54

$x=0,\dfrac{-3}{2},4,\dfrac{-5}{3}$

Given: $-7x[x(2x-5)-12](9x^2+30x+25)=0$ Need to use factorization to solve the quadratic equation. This can be factorized as follows: $-7x(2x+3)(x-4)(3x+5)^2=0$ Now $-7x=0 \implies x=0$ and $(2x+3)=0$ This implies $x=\dfrac{-3}{2}$ and $(x-4)=0 \implies x=4$ and $(3x+5)^2=0$ This implies $x=\dfrac{-5}{3}$ Hence, $x=0,\dfrac{-3}{2},4,\dfrac{-5}{3}$