Question

Factorise : $a^2-10a+25$

Answer 1

Given: $a^2-10a+25$

$=a^2-5a-5a+25$, split middle term so that resultant is same

$=(a^2-5a)+(-5a+25)$, group pair of terms

$=a(a-5)-5(a-5)$, factor each binomials

$=(a-5)(a-5)$, factor out common factor $(a-5)$

$=(a-5{)}^2$

Aliter, use $(a-b^2)=a^2-2ab+b^2$

$\Rightarrow a^2-10a+25=a^2-2\times 5x+5^2=(a-5{)}^2$