Question

(n-10)^2+(10-n) with process

Answer 1

$\textbf{Answer :}$

Now,

(n - 10)^2 + (10 - n)

= (n - 10)^2 - (n - 10)

= (n - 10) (n - 10 - 1)

= (n - 10) (n - 11),

which is the required factorization.

#$\textbf{MarkAsBrainliest}$

Answer 2

${(n - 10)}^{2} + 10 - n = 0$

${n}^{2} - 20n + 100 + 10 - n = 0$

${n}^{2} - 21n + 110 = 0$

${n}^{2} - 11n - 10n + 110 = 0$

$n(n - 11) - 10(n - 11) = 0$

$(n - 11)(n - 10) = 0$

$n - 11 = 0$

$n - 10 = 0$

$n = 10 \: \: \: \: and \: \: \: n = 11$