Question

If x² - 1 is a factor of ax⁴+bx³+cx²+dx+e , show that a+c+e = b+d =0 ​

Answer 1

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

Given that

$\rm :\longmapsto\: {x}^{2} - 1 \: is \: factor \: of \: a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e$

Let assume that

$\rm :\longmapsto\: f(x) \: = \: a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e$

Since,

$\rm :\longmapsto\: {x}^{2} - 1 \: is \: factor \: of \: f(x)$

$\rm :\longmapsto\: ({x}- 1)(x + 1) \: is \: factor \: of \: f(x)$

So,

Two cases arises.

Case :- 1

$\rm :\longmapsto\: ({x}- 1) \: is \: factor \: of \: f(x)$

We know,

Factor theorem :- It states that if (x - a) is a factor of polynomial f(x) then f(a) = 0.

So, by using Factor theorem,

$\rm :\longmapsto\:f(1) = 0$

$\rm :\longmapsto\: \: a {(1)}^{4} + b {(1)}^{3} + c {(1)}^{2} + d(1) + e = 0$

$\rm :\longmapsto\: \: a + b + c + d + e = 0 - - - (1)$

Case :- 2

$\rm :\longmapsto\: ({x} + 1) \: is \: factor \: of \: f(x)$

So, by Factor theorem,

$\rm :\longmapsto\:f( - 1) = 0$

$\rm :\longmapsto\: \: a {( - 1)}^{4} + b {( - 1)}^{3} + c {( - 1)}^{2} + d( - 1) + e = 0$

$\rm :\longmapsto\: \: a - b + c - d + e = 0 - - - (2)$

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

$\rm :\longmapsto\: \: 2(a + c + e) = 0$

$\rm :\longmapsto\: \: a + c + e= 0 - - - - (3)$

On substituting equation (3) in equation (1), we get

$\rm :\longmapsto\: \: b + d = 0 - - - (4)$

So, On combining equation (3) and (4), we get

$\rm :\longmapsto\: \: b + d =a + c + e = 0$

Hence, Proved

Additional Information:-

Remainder Theorem :-

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