Question

If both $x+2$ and $2x+1$ are factors of $a{x}^2+2x+b$, then the value of $a-b$?

Answer 1

Step 1. Compare the quadratic equation with standard form

We know that if $\alpha ,\beta$ are the roots of a quadratic equation $p{x}^2+qx+r=0$ then

1. $\alpha +\beta =-\frac{q}{p}$
2. $\alpha \beta =\frac{r}{p}$

Now comparing this with the given quadratic equation $a{x}^2+2x+b$ we get

1. $p=a$
2. $r=b$

Now $x+2$ and $2x+1$ are the factors of the polynomial .

$$\begin{gathered} \left(x+2\right)\left(2x+1\right)=0 \\ \therefore \left(x+2\right)=0 \\ \Rightarrow x=-2 \\ \left(2x+1\right)=0 \\ \Rightarrow x=-\frac{1}{2} \end{gathered}$$

$\therefore$ The roots of the polynomial are $-2$and $-\frac{1}{2}$

Step 2. Find $a-b$

Let $\alpha =2$ and $\beta =\frac{1}{2}$

$$\begin{gathered} \therefore \frac{r}{p}=\alpha \beta \\ \Rightarrow \frac{r}{p}=(-2)·(-\frac{1}{2}) \\ \Rightarrow \frac{b}{a}=1\left[\because r=b\mathrm{and}\mathrm{p}=\mathrm{a}\right] \\ \Rightarrow b=a \\ \Rightarrow a-b=0 \end{gathered}$$

Hence, value of $a-b$ is $0$.