Question

If $(x+3)$ and $(x+\frac{1}{3})$ are factors of $a{x}^2+bx+5$, then the value of $10a–3b+1$ is equal to

• A. 1
• B. 8
• C. 4
• D. -2

Answer 1

The correct option is A 1

Let $f\left(x\right)=a{x}^2+bx+5$
It is given that $(x+3)$ and $x+\frac{1}{3}$ are factors of f(x).
By factor theorem, $f(-3)=0$ and $f\left(\frac{-1}{3}\right)=0$
$\therefore f(–3)=a(–3{)}^2+b(–3)+5=0$
$\Rightarrow 9a-3b+5=0$.........(i)
$f\left(\frac{-1}{3}\right)=a(\frac{-1}{3}{)}^2+b(\frac{-1}{3})+5=0$
$\Rightarrow \frac{a}{9}-\frac{b}{3}+5=0$
$\Rightarrow a-3b+45=0$...........(ii)
Subtracting (ii) from (i), we get:
8a – 40 = 0
$\Rightarrow a=5$ …..(iii)
Substituting a = 5 in (i), we get:
45 – 3b + 5 = 0
$\Rightarrow 3b=50$
$\Rightarrow b=\frac{50}{3}$.........(iv)
Using (iii) and (iv), we get:
10a – 3b + 1
$=10\left(5\right)-3\left(\frac{50}{3}\right)+1$
= 50 -50 + 1
= 1
Hence, the correct answer is option (1).