If $(x + l) (x + m) = x^{2} + 4 x + 2$ find $l^{2} + m^{2}$ and $(l - m)^{2}$

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Solution

By product formula, we know
$(x + l) (x + m) = x^{2} + (l + m) x + l m$
So, by comparing RHS with $x^{2} + 4 x + 2$ , we have
$l + m = 4$ and $l m = 2$
Now, $l^{2} + m^{2} = (l + m)^{2} - 2 l m$
$= 4^{2} - 2 (2) = 16 - 4$
$l^{2} + m^{2} = 12$
$(l - m)^{2} = (l + m)^{2} - 4 l m$
$= 4^{2} - 4 (2) = 16 - 8$
$(l - m)^{2} = 8$

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