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)
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.
Please work the complete solutions.
3 answers
3B)
ax^2 + bx + c = a (x-alpha) (x-beta)
Evaluate the derivative of the square of both sides at x = 0.
ax^2 + bx + c = a (x-alpha) (x-beta)
Evaluate the derivative of the square of both sides at x = 0.
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.
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.
2 answers