Question

a (x^2 + 1) - x (a^2 +1)
Find zeros of polynomial

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:p(x) = a( {x}^{2} + 1) - x( {a}^{2} + 1)$

$\rm :\longmapsto\:p(x) = a{x}^{2} +a - x{a}^{2} - x$

$\rm :\longmapsto\:p(x) = a{x}^{2} - x{a}^{2} + a - x$

$\rm :\longmapsto\:p(x) = (a{x}^{2} - x{a}^{2}) + (a - x)$

$\rm :\longmapsto\:p(x) = ax({x} - {a}) + (a - x)$

$\rm :\longmapsto\:p(x) = ax({x} - {a}) - 1 (x - a)$

$\rm :\longmapsto\:p(x) = (ax - 1)({x} - {a})$

Now, zeroes of polynomial p(x) means those real values of x for which p(x) = 0.

So, to find zeroes of p(x),

$\rm :\longmapsto\: (ax - 1)({x} - {a}) = 0$

$\rm :\longmapsto\: ax - 1 = 0 \: \: or \: \: {x} - {a}= 0$

$\rm :\longmapsto\: ax= 1\: \: or \: \: {x}= 0a$

$\bf\implies \:x = \: \dfrac{1}{a} \: \: \: or \: \: \: x = a$

So, Zeroes of

$\red{\boxed{ \bf{ \: \rm :\longmapsto\: a( {x}^{2} + 1) - x( {a}^{2} + 1) \: are \: a \: and \: \frac{1}{a} }}}$

Additional Information :-

Let solve one more problem!!

Question :- Find the zeroes of the following

$\rm :\longmapsto\: {abx}^{2} + ( {b}^{2} - ac)x - cb$

Solution :-

Let assume that

$\rm :\longmapsto\: p(x) = {abx}^{2} + ( {b}^{2} - ac)x - cb$

$\rm :\longmapsto\: p(x) = {abx}^{2} + {b}^{2}x - acx - cb$

$\rm :\longmapsto\: p(x) = {bx}(ax + {b}) - c(ax + b)$

$\rm :\longmapsto\: p(x) = (ax + {b})(bx - c)$

To find zeroes of p(x), we have

$\rm :\longmapsto\:(ax + {b})(bx - c) = 0$

$\rm :\longmapsto\:ax + {b} = 0 \: \: or \: \: bx - c = 0$

$\rm :\longmapsto\:ax= - b \: \: or \: \: bx = c$

$\bf\implies \:x = - \dfrac{b}{a} \: \: \: or \: \: \: x = \dfrac{c}{b}$

So, zeroes of

$\red{\boxed{ \bf{ \: \rm :\longmapsto\: {abx}^{2} + ( {b}^{2} - ac)x - cb \: are \: - \frac{b}{a} \: and \: \frac{c}{b}}}}$