Question
If $x^2 - 1$ is a factor of $ax^4 +bx ^3 + cx^2 + dx + e,$ then.
✓
Answer
As$ (x^2 - 1)$ Is a factor of polynomial
$f(x^2) = ax^4 + bx^3 + cx^2 + dx + e$
Therefore$,$
$f(x) = 0$
And,
$f(1) = 0$
$a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = 0$
$\Rightarrow a + b + c + d + e = 0$
And,
$f(-1) = 0$
$a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) + e = 0$
$a - b - c - d + e = 0$
Hence$, a + c + e = b + d$
$f(x^2) = ax^4 + bx^3 + cx^2 + dx + e$
Therefore$,$
$f(x) = 0$
And,
$f(1) = 0$
$a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = 0$
$\Rightarrow a + b + c + d + e = 0$
And,
$f(-1) = 0$
$a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) + e = 0$
$a - b - c - d + e = 0$
Hence$, a + c + e = b + d$
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