Question 582164
{{{ a^2 + 16a + 15 = ( a + 15 )*( a + 1 ) }}}
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You can find this by completing the square
{{{ a^2 + 16a + (16/2)^2 = -15 + (16/2)^2 }}}
{{{ a^2 + 16a + 64 = -15 + 64 }}}
{{{ ( a + 8 )^2 = 49 }}}
{{{ ( a + 8 )^2 = 7^2 }}}
{{{ a + 8 = 7 }}}
{{{ a = -1 }}}
and, also,
{{{ a + 8 = -7 }}}
{{{ a = -15 }}}
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rewrite these as:
{{{ a + 1 = 0 }}}
{{{ a + 15 = 0 }}}
and
{{{ ( a + 1 )*( a + 15 ) = 0 }}}
{{{ ( a + 1 )*( a + 15 ) = a^2 + 16a + 15 }}}