Question

I have two of these factoring questions on my Algebra II homework and need further explanation. Please help!

Answer 1

`10x^(4k+6)-7x^(2k+3)-12\qquad \begin{cases} 4k+6=2(2k+3)\\\\ \stackrel{\textit{let's call}}{2k+3=u} \end{cases}\implies 10x^(2u)-7x^u-12 \\\\\\ 10(x^u)^2-7(x^u)-12\implies ( ~~2x^u-3 ~~ )( ~~5x^u+4 ~~ ) \\\\[-0.35em] ~\dotfill\\\\ ~\hspace{6em} \stackrel{\textit{substituting back}}{ {\Large \begin{array}{llll} ( ~~2x^(2k+3)-3 ~~ )( ~~5x^(2k+3)+4 ~~ ) \end{array}}}~\hfill \\\\[-0.35em] \rule{34em}{0.25pt}`

`20d^(2r+16)-23d^(r+8)+6\qquad \begin{cases} 2r+16=2(r+8)\\\\ \stackrel{\textit{let's call}}{r+8=u} \end{cases}\implies 20d^(2u)-23d^u+6 \\\\\\ 20(d^u)^2-23(d^u)+6\implies ( ~~ 4d^u-3 ~~ )( ~~ 5d^u-2 ~~ ) \\\\[-0.35em] ~\dotfill\\\\ ~\hspace{6em} \stackrel{\textit{substituting back}}{ {\Large \begin{array}{llll} ( ~~ 4d^(r+8)-3 ~~ )( ~~ 5d^(r+8)-2 ~~ ) \end{array}}}`