Question

If $(x-3)$ and $\left(x-\frac{1}{3}\right)$ are both factors of $a{x}^2+5x+b$, show that $a=b$.

Answer 1

If $(x-3)$ is a factor of $f\left(x\right)=a{x}^2+5x+b$ then by factor theorem,

$\Rightarrow f\left(3\right)=0$

$\Rightarrow f\left(3\right)=a(3{)}^2+5(3)+b=0$

$\Rightarrow 9a+15+b=0$ ...(i)

If $(x-\frac{1}{3})$ is a factor of $f\left(x\right)=a{x}^2+5x+b$ then by factor theorem,

$\Rightarrow f\left(\frac{1}{3}\right)=0$

$\Rightarrow f\left(\frac{1}{3}\right)=a(\frac{1}{3}{)}^2+5(\frac{1}{3})+b=0$

$\Rightarrow \frac{a}{9}+\frac{5}{3}+b=0$

$\Rightarrow a+15+9b=0$...(ii)

From (i) and (ii),

$\Rightarrow 9a+15+b=a+15+9b$

$\Rightarrow 8a=8b$

$\therefore a=b$