Let f(x) = 2×3 – 3×2 – 17x + 30 be the given polynomial. The factors of the constant term +30 are {tex}\\pm 1, \\pm 2, \\pm 3, \\pm 5, \\pm 6, \\pm 10, \\pm 15, \\pm 30{/tex}. The factor of coefficient of x3 is 2. Hence, possible rational roots of f(x) are:{tex}\\pm 1, \\pm 3, \\pm 5, \\pm 15, \\pm \\frac{1}{2}, \\pm \\frac{3}{2}, \\pm \\frac{5}{2}, \\pm \\frac{{15}}{2}{/tex}We have f(2) = 2(2)3 – 3(2)2 – 17(2) + 30= 2(8) – 3(4) – 17(2) + 30= 16 – 12 – 34 + 30 = 0And f(-3) = 2(-3)3 – 3(-3)2 – 17(-3) + 30= 2(-27) – 3(9) – 17(-3) + 30= -54 – 27 + 51 + 30 = 0So, (x – 2) and (x + 3) are factors of f(x).{tex}\\Rightarrow{/tex} x2 + x – 6 is a factor of f(x).Let us noe divide f(x) = 2×3 – 3×2 – 17x + 30 by x2 + x – 6 to get the other factors of f(x).Factors of f(x).By long division, we have{tex}\\therefore{/tex} 2×3 – 3×2 – 17x + 30 = (x2 + x – 6)(2x – 5){tex}\\Rightarrow{/tex} 2×3 – 3×2 – 17x + 30 = (x – 2)(x + 3)(2x – 5)Hence, 2×3 – 3×2 – 17x + 30 = (x – 2)(x + 3)(2x – 5)