Question:
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) $a=5, b=-1$
Solution:
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
$4=4 b$
$\frac{4}{4}=b$
$1=b$
By substituting $b=1$ in $a+b=4$ we get
$a+1=4$
$a=4-1$
$a=3$
Then $a=3, b=1$
Hence, the correct choice is $(b)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.