MCQ
If $x^2-1$ is a factor of $a x^4+b x^3+c x^2+d x+e$, then
- ✓$a + c + e = b + d$
- B$a + b + e = c + d$
- C$a + b + c = d + e$
- D$b + c + d = a + e$
✓
Answer
Correct option: A.
$a + c + e = b + d$
If $x^2-1$ is factor of $p(x)=a x^4+b x^3+c x^2+d x+e$.
Then $(x - 1)$ and $(x + 1)$ will also be factors of $p(x).$
Because $x^2 - 1 = (x - 1)(x + 1)$
Then, at $x = 1$ and $x = -1, p(x) = 0$
$⇒ p(1) = 0$ and $p(-1) = 0$
$⇒ a + b + c + d + e = 0 ...(1)$
And
$⇒ a - b + c - d + e = 0 ...(2)$
Adding equations $(1)$ and $(2).$
$2a + 2c + 2e = 0$
$⇒ a + c + e = 0 ...(3)$
Substracting equation $(2)$ from $(1)$
$2b + 2d = 0$
$⇒ b + d = 0 ...(4)$
From equations $(3)$ and $(4),$ we get
$a + c + e = b + d$
Then $(x - 1)$ and $(x + 1)$ will also be factors of $p(x).$
Because $x^2 - 1 = (x - 1)(x + 1)$
Then, at $x = 1$ and $x = -1, p(x) = 0$
$⇒ p(1) = 0$ and $p(-1) = 0$
$⇒ a + b + c + d + e = 0 ...(1)$
And
$⇒ a - b + c - d + e = 0 ...(2)$
Adding equations $(1)$ and $(2).$
$2a + 2c + 2e = 0$
$⇒ a + c + e = 0 ...(3)$
Substracting equation $(2)$ from $(1)$
$2b + 2d = 0$
$⇒ b + d = 0 ...(4)$
From equations $(3)$ and $(4),$ we get
$a + c + e = b + d$
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