Question 363908
{{{3x^2+12x+5=p(x+q)^2+r}}} Start with the given equation.



{{{3x^2+12x+5=p(x^2+2qx+q^2)+r}}} FOIL




{{{3x^2+12x+5=px^2+2pqx+pq^2+r}}} Distribute.



Notice that the coefficients for the {{{x^2}}} terms are {{{3}}} (on the left) and {{{p}}} (on the right). So {{{p=3}}}



So the equation becomes {{{3x^2+12x+5=3x^2+2*3qx+3q^2+r}}}



{{{3x^2+12x+5=3x^2+6qx+3q^2+r}}} Multiply



Now notice that the coefficients for the x terms are 12 and 6q. So {{{6q=12}}}, which means that {{{q=2}}}



So the equation then becomes {{{3x^2+12x+5=3x^2+6*2x+3(2)^2+r}}} and simplifies to {{{3x^2+12x+5=3x^2+12x+12+r}}}



Finally, we see that 12+r=5, which means that {{{r=-7}}}