If x + 2 is a factor of x2 + ax + 2b and a + b = 4,


If $x+2$ is a factor of $x^{2}+a x+2 b$ and $a+b=4$, then

(a) $a=1, b=3$

(b) $a=3, b=1$

(c) $a=-1, b=5$

(d) $a=5, b=-1$


Given that $x+2$ is a factor of $x^{2}+a x+2 b$ and $a+b=4$

$f(x)=x^{2}+a x+2 b$

$f(-2)=(-2)^{2}+a(-2)+2 b$

$0=4-2 a+2 b$

$-4=-2 a+2 b$

By solving $-4=-2 a+2 b$ and $a+b=4$ by elimination method we get

Multiply $a+b=4$ by 2 we get,

$2 a+2 b=8$. So

$4=4 b$



By substituting $b=1$ in $a+b=4$ we get




Then $a=3, b=1$

Hence, the correct choice is $(b)$

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