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Given a polynomial P(x) along degree at least 1 & any number r there is another polynomial Q(x), called as the quotient, with degree one less than degree of P(x) & a number R, called the remainder, such that,
P ( x ) = ( x - r ) Q ( x ) + R
Note as well that Q(x) and R is unique, or in other terms, there is only one Q(x) & R that will work for a given P(x) & r.
Thus, with the one example we've done to this point we can illustrates that,
Q ( x ) = 5x2 + 19 x + 76 and R = 310
Given f ( x ) = x 2 - 2 x + 8 and g( x ) = √(x+ 6) evaluate f (3) and g(3) Solution Okay we've two function evaluations to do here and we've also obtained two functions
what are some facts about composition of functions?
BEST WAY TO FIND ORDERED PAIRS
Any vector space V satisÖes the ten axioms, among which the last one is: "for any vector * u 2 V; 1 * u = * u; where 1 is the multiplicative identity of real numbers R:" Discuss th
Whereas we are on the subject of function evaluation we have to now talk about piecewise functions. Actually we've already seen an instance of a piecewise function even if we didn'
Find out the partial fraction decomposition of each of the following. 8x 2 -12/( x( x 2 + 2 x - 6) Solution In this case the x which sits in the front is a linear term
w^2 + 30w + 81= (-9x^3 + 3x^2 - 15x)/(-3x) (14y = 8y^2 + y^3 + 12)/(6 + y) ac + xc + aw^2 + xw^2 10a^2- 27ab + 5b^2 For the last problem I have to incorporate the following words
In previous section we looked at the two functions f ( x) = 3x - 2 and g ( x )= x/3 + 2/3 and saw that ( f o g ) ( x ) =(g o f )( x ) =
Properties of Logarithms 1. log b 1 = 0 . It follows from the fact that b o = 1. 2. log b b = 1. It follows from the fact that b 1 = b . 3. log b b x = x . it c
In the last two sections of this chapter we desire to discuss solving equations & inequalities that have absolute values. We will look at equations along with absolute value in th
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