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IIf x^2-1 is a factor of ax^4+bx^3+cx^2+dx+e. Show that a+c+e=b+d=0?
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IIf x^2-1 is a factor of ax^4+bx^3+cx^2+dx+e. Show that a+c+e=b+d=0?
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IIf x^2-1 is a factor of ax^4+bx^3+cx^2+dx+e. Show that a+c+e=b+d=0?

Proof:

Given:
x^2-1 is a factor of ax^4 + bx^3 + cx^2 + dx + e

Expression:
ax^4 + bx^3 + cx^2 + dx + e = (x^2 - 1)(px^2 + qx + r)

Expanding the RHS:
px^4 + qx^3 + rx^2 - px^2 - qx - r = ax^4 + bx^3 + cx^2 + dx + e

Equating coefficients of x^4, x^3, x^2, x, and the constant term:
a = p
b = q - p
c = r - p
d = -q
e = -r

Substitute the values of p, q, and r:
a + c + e = a + (r - p) + e = p + (r - p) + e = r + e = 0
b + d = (q - p) - q = -p = 0

Therefore, a + c + e = 0 and b + d = 0 as required.
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IIf x^2-1 is a factor of ax^4+bx^3+cx^2+dx+e. Show that a+c+e=b+d=0?
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