Question

pls tell answer of 16th and 19th​

Answer 1

Answer:

16)

$(\sqrt{5}x+4 )(5x+2\sqrt{5} )$

17)

$p(p-q)(p+q)(p^{2} +q^{2})$

Step-by-step explanation:

16)

$5\sqrt{5}x^{2} +30x+8\sqrt{5}\\\\=> 5\sqrt{5} x^{2} +20x+10x+8\sqrt{5}\\\\=>5\sqrt{5}x^{2} +4\times5x+2\sqrt{5}\times\sqrt{5}x+8\sqrt{5} \\\\=>5x(\sqrt{5}x+4 )+2\sqrt{5} (\sqrt{5}x+4 )\\\\=>(\sqrt{5}x+4 )(5x+2\sqrt{5} )$

17)

$p^{7}-pq^{6}\\\\=>p(p^{6}-q^{6} )\\\\=>p[(p^{3})^{2}-(q^{3})^{2}]\\\\=>p[(p^{2} -q^{2} )(p^{2} +q^{2} )] \\ \ \ \ using \ Identity \ (a^{2}-b^{2})=(a+b)(a-b) \\=>p(p^{2} -q^{2} )(p^{2} +q^{2} )\\\\=>p[(p)^{2} -(q)^{2}](p^{2} +q^{2} )\\\\=>p[(p-q)(p+q)](p^{2} +q^{2})\\\\=>p(p-q)(p+q)(p^{2} +q^{2})$

Answer 2

Answer:

$using Identity (a2−b2)=(a+b)(a−b)=>p(p2−q2)(p2+q2)=>p[(p)2−(q)2](p2+q2)=>p[(p−q)(p+q)](p2+q2)=>p(p−q)(p+q)(p2+q2)$