Question

The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)

Answer 1

$\bf ~\hspace{10em}\textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 8x^3-125~~ \begin{cases} 8=2\cdot 2\cdot 2\\ \qquad 2^3\\ 125=5\cdot 5\cdot 5\\ \qquad 5^3 \end{cases}\implies 2^3x^3-5^3\implies (2x)^3-5^3 \\[2em] [2x-5][(2x)^2+(2x)(5)+5^2]\implies (2x-5)(\stackrel{\stackrel{j}{\downarrow }}{4}x^2+\stackrel{\stackrel{k}{\downarrow }}{10}x+25)$