Question 1152320
<pre>

{{{ (x+2)^2 + b(x+2) + a = (x+6)(x+1) }}}
Expand both sides:
{{{ (x^2+4x+4) + (bx+2b) + a =  x^2+7x+6 }}}

Collect like terms on LHS:
{{{ (x^2) + (b+4)x + (2b+a+4) =  x^2+7x+6 }}}

Now compare coefficents of {{{x^2}}},{{{x^1}}}, {{{x^0}}} terms from LHS and RHS:

{{{ b+4 = 7 }}}
{{{ 2b+a+4 = 6 }}}

Solve for a and b:  {{{ highlight( a=-4 )}}}, {{{highlight( b=3 )}}}

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Check:
{{{ x^2+4x+4 + 3(x+2) + (-4) }}} =
{{{ x^2+4x+4 + 3x+6 -4 }}} = 
{{{ x^2+7x+6 }}} =
{{{ (x+6)(x+1) }}}  (ok)