Question

if both x-2 and x-1/2 are factors of px^2+5x+r,prove p=r​

Answer 1

Basic Concept Used :-

Factor Theorem :-

This theorem states that, if x - a, is a factor of polynomial f(x), then f(a) = 0.

Solution :-

$\rm :\longmapsto\:Let \: f(x) = {px}^{2} + 5x + r$

Given that,

$\rm :\longmapsto\:x - 2 \: is \: a \: factor \: of \: f(x) = {px}^{2} + 5x + r$

So,

$\rm :\longmapsto\:f(2) = 0$

$\rm :\longmapsto\:p {(2)}^{2} + 5 \times 2 + r = 0$

$\rm :\longmapsto\:4p + 10 + r = 0$

$\rm :\longmapsto\:4p + r = - \: 10 - - - (1)$

Also,

$\rm :\longmapsto\:x - \dfrac{1}{2} \: is \: a \: factor \: of \: f(x) = {px}^{2} + 5x + r$

$\rm :\longmapsto\:f \bigg( \dfrac{1}{2} \bigg) = 0$

$\rm :\longmapsto\:p {\bigg(\dfrac{1}{2} \bigg) }^{2} + 5 \times \dfrac{1}{2} + r = 0$

$\rm :\longmapsto\:p {\bigg(\dfrac{1}{4} \bigg) } + \dfrac{5}{2} + r = 0$

$\rm :\longmapsto\:\dfrac{p + 10 + 4r}{4} = 0$

$\rm :\longmapsto\:p + 4r = - 10 - - - (1)$

Now,

On equating equation (1) and equation (2), we get

$\rm :\longmapsto\:p + 4r = 4p + r$

$\rm :\longmapsto\:3p = 3r$

$\rm :\implies\:p = r$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information

Remainder Theorem :-

This theorem states that if polynomial f(x) is divided by x - a, then remainder is f(a).