Question

# If x + 2 is a factor of ${{x}^{2}}+ax+2b=0$ and a + b = 4, thenA.a = 1, b = 3B.a = 3, b = 1C.a = -1, b = 5D.a = 5, b = -1

Hint: In this problem, we are required to find a and b. So, we need two equations to solve for a and b. One of the equations is given as a + b = 4. Another equation is formed by using the property of factor as factor divides the number completely. By putting x = -2, we get the second equation and our final values are obtained by substitution.

$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
A factor is an expression which divides the whole expression completely without leaving remainder. According to our problem, we are given that x + 2 is a factor of ${{x}^{2}}+ax+2b=0$. So, replacing x = -2, we get
\begin{align} & f(x)={{x}^{2}}+ax+2b \\ & f\left( -2 \right)={{\left( -2 \right)}^{2}}+a\left( -2 \right)+2b \\ & f\left( -2 \right)=4-2a+2b \\ & 4-2a+2b=0 \\ & 2b=2a-4 \\ & b=a-2\ldots (1) \\ \end{align}
\begin{align} & a+a-2=4 \\ & 2a=4+2 \\ & a=\dfrac{6}{2} \\ & a=3 \\ & b=3-2 \\ & b=1 \\ \end{align}