Question

factorise :- 16p² - 289q²​

Answer 1

Answer:

${(4p ){}^{2} - (17q ){}^{2} } =( 4p + 17q) \times (4p - 17q)$

Answer 2

${16p}^{2} - {289q}^{2}$

=

${(4p)}^{2} - {(17q)}^{2}$

=

$(4p + 17q)(4p - 17q)$

In third step there is identity of

${a}^{2} - {b}^{2} = (a + b)(a - b)$

$<marquee >This is your answer$

Thank You