Question 164436
{{{a^2 - 13a + 40}}}
I can set the equation equal to zero
and complete the square to find roots
{{{a^2 - 13a + (-13/2)^2 = -40 + (-13/2)^2}}}
{{{a^2 - 13a + 169/4 = -160/4 + 169/4}}}
{{{(a - 13/2)^2 = 9/4}}}
Take the square root of both sides
{{{a - 13/2 = 3/2}}}
{{{2a - 13 = 3}}}
{{{2a = 16}}}
{{{a = 8}}}
{{{a - 8 = 0}}}
and, taking the negative square root
{{{a - 13/2 = -(3/2)}}}
{{{2a - 13 = -3}}}
{{{2a = 10}}}
{{{a = 5}}}
{{{a - 5 = 0}}}
{{{a^2 - 13a + 40 = (a-8)(a-5)}}}