X^5 + ax^3 + bx^2 - 3 = (x^2 - 1)Q(x) - x - 2

find the value of a and b?

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Step-by-step explanation:

Subtracting the remainder from the original polynomial gives a polynomial that has (x^2 -1) as a factor. Dividing, we find the quotient and remainder to be ...

  (x^5 +ax^3 +bx^2 +x -1)/(x^2 -1) = (x^3 +(1+a)x +b) remainder (2+a)x +(b -1)

Since we want the remainder to be zero, we require ...

  2+a = 0   ⇒   a = -2

  b -1 = 0   ⇒   b = 1

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The attachment shows the division as given in the problem statement (with the given remainder expression).

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: