Thread: Factorization

Factorize:
25- p^2 - q^2 - 2pq is
(a)(5+p+q)(5-p+q)
(b)(5+p+q)(5-p-q)
(c)(5+p+q)(5+p-q)
(d)(5+p-q)(5-p+q)

Hello,

Hints: (p+q)^2=? (p-q)^2=? (p+q)(p-q)=?

Warnig: Don' t forget the parentheses for grouping!

Bye.

Hello, devi!

Factorize: .$\displaystyle 25- p^2 - q^2 - 2pq$

. . $\displaystyle \begin{array}{cc}(a)\;(5+p+q)(5-p+q) & (b)\;(5+p+q)(5-p-q) \\ \\[-3mm] (c)\;(5+p+q)(5+p-q) & (d)\;(5+p-q)(5-p+q)\end{array}$

We have: .$\displaystyle 25 - (p^2 + 2pq + q^2) \;=\;25 - (p + q)^2\quad\hdots$ a difference of squares!

. . . $\displaystyle =\;\bigg[5 + (p+q)\bigg]\,\bigg[5 - (p+q)\bigg] \;=\;(5 + p + q)(5 - p - q)\;\;{\color{blue}(b)}$