Answer
$128x+26x\sqrt{14}$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$, the given expression, $
\left( 5\sqrt{7x}-\sqrt{2x} \right)\left( 4\sqrt{7x}+6\sqrt{2x} \right)
,$ is equivalent to
\begin{array}{l}\require{cancel}
5\sqrt{7x}(4\sqrt{7x})+5\sqrt{7x}(6\sqrt{2x})-\sqrt{2x}( 4\sqrt{7x})-\sqrt{2x}(6\sqrt{2x})
.\end{array}
Using the properties of radicals, the expression above simplifies to
\begin{array}{l}\require{cancel}
5(4)\sqrt{7x(7x)}+5(6)\sqrt{7x(2x)}-4\sqrt{2x(7x)}-6\sqrt{2x(2x)}
\\\\=
20\sqrt{(7x)^2}+30\sqrt{14x^2}-4\sqrt{14x^2}-6\sqrt{(2x)^2}
\\\\=
20\sqrt{(7x)^2}+30\sqrt{x^2\cdot 14}-4\sqrt{x^2\cdot 14}-6\sqrt{(2x)^2}
\\\\=
20\sqrt{(7x)^2}+30\sqrt{(x)^2\cdot 14}-4\sqrt{(x)^2\cdot 14}-6\sqrt{(2x)^2}
\\\\=
20(7x)+30(x)\sqrt{14}-4(x)\sqrt{14}-6(2x)
\\\\=
140x+30x\sqrt{14}-4x\sqrt{14}-12x
\\\\=
128x+26x\sqrt{14}
.\end{array}
Note that the variables are assumed to have positive values.
