Question

find the product using identity
(p²+16) (p²-¼) ​

Answer 1

Given :

• $( p^2 + 16 ) ( p^2 - \frac{1}{4} )$

To find :

• Product of $( p^2 + 16 ) ( p^2 - \frac{1}{4} )$

According to the question :

⟹ $( p^2 + 16 ) ( p^2 - \frac{1}{4} )$

⟹ $[ p^2 × ( p^2 - \frac{1}{4} ) ] + [ 16 × ( p^2 - \frac{1}{4} )$

⟹ $[ p^{( 2 + 2 )} - ( p^2 × 1 / 4 ) ] + [ ( 16 × p^2 ) - ( 16 × \frac{1}{4} ) ]$

⟹ $p^4 - \frac{1}{4}p^2 + 16p^2 - \frac{16}{4}$

⟹ $p^4 - \frac{1}{4}p^2 + 16p^2 - 4$

⟹ $p^4 + \frac{[ ( 16 × 4 ) - 1 ]}{4} p^2 - 4$

⟹ $p^4 + \frac{[ ( 64 - 1 ) p^2 ]}{4} - 4$

⟹ $\bold {p^4 + \frac{63}{4} p^2 - 4}$

So, It's Done !!

Answer 2

Given:

$\sf \bull ( {p}^{2} + 16)( {p}^{2} - \frac{1}{4} )$

Find:

$\sf \bull solve \: using \: suitable \: identity$

Solution:

we, have

$\sf \to ( {p}^{2} + 16)( {p}^{2} - \frac{1}{4} )$

$\sf by \: using \: (x + a)(x + b) = {x}^{2} + (a + b)x + ab$

So,

$\sf : \implies ( {p}^{2} + 16)( {p}^{2} - \frac{1}{4} )$

$\sf : \implies { ({p}^{2}) }^{2} + (16+ ( \frac{ - 1}{4} )) {p}^{2} + (16) \times ( \frac{ - 1}{4} )$

$\sf : \implies {p }^{4} + (16 - \frac{ 1}{4}) {p}^{2} + ( \frac{ - 16}{4} )$

$\sf : \implies {p }^{4} + (\frac{64 - 1}{4}) {p}^{2} - \frac{ 16}{4}$

$\sf : \implies {p }^{4} + (\frac{63}{4}) {p}^{2} - \frac{ 16}{4}$

$\sf : \implies {p }^{4} + \dfrac{63}{4}{p}^{2} - \dfrac{ 16}{4}$

$\sf : \implies \dfrac{4{p }^{4} + 63{p}^{2} - 16}{4}$

$\sf : \implies \dfrac{4{p}^{4}}{4} + \dfrac{63{p}^{2} }{4} - \dfrac{16}{4}$

$\sf : \implies {p}^{4}+ \dfrac{63{p}^{2} }{4} - 4$

________________

Hence, the answer will be $\sf {p}^{4}+ \dfrac{63{p}^{2} }{4} - 4$