Question

If $x^{2}$ −1 is a factor of $ax^{4}-bx^{3}+cx^{2} +dx+e$, then find the value of a + c + e .

Answer 1

$\huge\red{brainliest\:plz}$ ❤️

Answer 2

   

Hey mate!!!

 

If x² - 1 is a factor of p(x) = ax⁴ - bx³ + cx² + dx + e, then 1 and -1 are two zeroes of p(x), because x² - 1 = (x - 1)(x + 1).

So that p(1) = p(-1) = 0.

$\Rightarrow\ p(1)=a(1)^4-b(1)^3+c(1)^2+d(1)+e=0 \\ \\ \Rightarrow\ p(1)=a-b+c+d+e=0 \\ \\ \\ \Rightarrow\ a-b+c+d+e=0 \\ \\ \Rightarrow\ a+c+e=b-d\ \ \ \ \ \longrightarrow\ \ \ \ \ (1)$

$\Rightarrow\ p(-1)=a(-1)^4-b(-1)^3+c(-1)^2+d(-1)+e=0 \\ \\ \Rightarrow\ p(-1)=a(1)-b(-1)+c(1)-d+e=0 \\ \\ \Rightarrow\ p(-1)=a+b+c-d+e=0 \\ \\ \\ \Rightarrow\ a+b+c-d+e=0 \\ \\ \Rightarrow\ a+c+e=d-b\ \ \ \ \ \longrightarrow\ \ \ \ \ (2)$

From (1) and (2),

$\Rightarrow\ b-d=d-b \\ \\ \Rightarrow\ b+b=d+d \\ \\ \Rightarrow\ 2b=2d \\ \\ \Rightarrow\ b=d \\ \\ \Rightarrow\ b-d=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{OR}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ d-b=0 \\ \\ \Rightarrow\ a+c+e=\bold{0}$

∴ a + c + e = 0

   

Plz ask me if you have any doubt on my answer.

Thank you...