Question

If both x - 2 and $x-\frac{1}{2}$ are factors of $p{x}^2$ + 5x + r, then

• A. p = r
• B. p + r = 0
• C. 2p + r = 0
• D. p + 2r = 0

Answer 1

The correct option is A

p = r

(x-2) and $(x-\frac{1}{2})$ are the factors of p(x) = $p{x}^2+5x+r$

Let x - 2 = 0, then x = 2

$\therefore f\left(2\right)=p(2{)}^2+5\times$ 2 + r = 4p + 10 + r = 4p + r + 10

$\therefore$ x - 2 is a factor of f(x)

$\therefore$ x - 2 is a factor of f(x) $\therefore$ Remainder = 0

$\therefore$ 4p + r + 10 = 0 $\Rightarrow$ 4p + r =-10 ...........(i)

Again let x - $\frac{1}{2}$ = 0, then x = $\frac{1}{2}$

$\therefore f\left(\frac{1}{2}\right)=p{\left(\frac{1}{2}\right)}^2+5\times \frac{1}{2}+r$

$=\frac{p}{4}+\frac{5}{2}$ + r

$\therefore x-\frac{1}{2}$ is a factor of f(x) $\therefore$ Remainder = 0

$\therefore \frac{p}{4}+\frac{5}{2}+r=0\Rightarrow \frac{p}{4}+r=\frac{-5}{2}$

$\Rightarrow$ p + 4r =- 10 .........(ii)

Subtracting (i) from (ii) - 3p + 3r = 0 $\Rightarrow$ 3p = 3r

$\Rightarrow$ p = r