# Math ( Polynomial )

posted by Jack

This time three questions - 1. If (x^2 - 1 ) is a factor of polynomial ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0.
2. Let R1 and R2 be the remainders when polynomials x^3 + 2x^2 - 5ax - 7 and x^ 3 + ax^2 - 12 x + 6 are divided by ( x + 1 ) and ( x - 2 ) respectively. If 2R1 + R2 = 6, find a.
3. If alpha and beta are the zeros of the polynomial ax^2 + bx + c then evaluateA. (alpha)^2 / beta + (beta)^2 / alpha
B. alpha^2 .beta + alpha.beta^2
C. 1/(alpha)^4 + 1/(beta)^4.

1. Count Iblis

3)

ax^2 + bx + c = a (x-alpha) (x-beta)

A) We can write:

alpha^2/beta + beta^2/alpha =

(alpha^3 + beta^3)/(alpha beta).

Putting x = 0 in the polynomial gives:

c = a alpha beta ---->

alpha beta = c/a

To find alpha^3 + beta^3, expand the logarithm of the polynomial in powers of 1/x:

Log(ax^2 + bx + c) =

Log(a) + Log(x-alpha) + Log(x-beta)

The expansion of the right hand side yields:

Log(a) + 2 Log(x) + Log(1-alpha/x) +
Log(1-beta/x) =

Log(a) + 2 Log(x) - [S1/x + 1/2 S2/x^2 + 1/3 S3/x^3 + ...]

where Sr = alpha^r + beta^r

To find S3 we thus need to expand
Log(ax^2 + bx + c) to order 1/x^3:

Log(ax^2 + bx + c) =

Log(a x^2) + Log[1+ b/(ax) + c/(ax^2)]=

Log(a x^2) + b/(ax) + c/(ax^2) +

- [b/(ax) + c/(ax^2)]^2/2 +

[b/(ax) + c/(ax^2)]^3/3

The coefficient of x^3 is thus:

b^3/(3a^3) - bc/a^2

and this is equal to -S3/3, therefore we have:

S3 = 3 bc/a^2 - b^3/a^3

Therefore:

(alpha^3 + beta^3)/(alpha beta) =

3 b/a - b^3/(c a^2)

2. Count Iblis

3B)

ax^2 + bx + c = a (x-alpha) (x-beta)

Evaluate the derivative of the square of both sides at x = 0.

3. Count Iblis

3C)

ax^2 + bx + c = a (x-alpha) (x-beta)

Expand the logarithm of both sides in powers of x, equate the coefficient of x^4 of both sides.

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