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POLYNOMIALS: It is not once nor twice but times without number that the same ideas make their appearance in the world.
1. Find the value for K for which x4 + 10x3 + 25x2 + 15x + K exactly divisible by x + 7.
Ans: Let P(x) = x4 + 10x4 + 25x2 + 15x + K and g(x) = x + 7
Since P(x) exactly divisible by g(x)
∴ r (x) = 0
(Ans : K= - 91)
now
x4 + 7 x3
-------------
3x3 + 25 x2
3x3 + 21x2
-------------------
4x2 + 15 x
4x2 + 28x
-----------------
-13x + K
- 13x - 91
----------------
K + 91
∴ K + 91 = 0
K= -91
Suppose S = {vi} and T = {ti} are "easy" sets of knapsak weight. Also, P and q are primes p > ?Si and q > ?ti. We can combine S and T into a signle set of knapsack weight as follow
the segments shown could form a triangle
BROKARAGE.
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We're here going to take a brief detour and notice solutions to non-constant coefficient, second order differential equations of the form. p (t) y′′ + q (t ) y′ + r (t ) y = 0
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