1. Simultaneous equations & equating coefficients: + B(x + 2) = Ax + 3A + Bx + 2B)
x + 3A + 2B = 4x + 9)
We can equate coefficients here and we get a pair of simultaneous equations:
&
:.
 = 9)
Substituting back in, you get
2. Calculus (or expansion if you like):I assume you meant
^3 + C)
, because otherwise this equation doesn't make alot of sense
This is the form of a cubic in point-of-inflection form, so using calculus to find this point:
Let
:
^2=0)
So we have
as a location for our point of inflection:
Substituting x = 3 into original equation:
so we have the pair:
)
Now remembering our general cubic form:
^3 + k)
, where h is the x-coordinate of the point of inflection and k is the y-coordinate. a is the dilation factor (in this case, it is 1, because in the LHS the
term has a coefficient of 1
so
^3 + 5 = x^3 - 9x^2 +27x - 22 )
:.
3: Polynomial Long Division: 
)
)
)
Remainder:
Remebering our form for long division:
}{d(x)} = q(x) + \dfrac{r(x)}{d(x)})
Where
 = x^4 - 9x^3 + 25x^2 - 8x - 2)
,
 = x^2 - 2)
,
 = 52 - 26x)
&
 = x^2 - 9x + 27)
:
 + \dfrac{52 - 26x}{x^2 - 2})
4. Inverse= 5 - \dfrac{2}{(x-6)^3})
This function is one-to-one, so it will have an inverse:
To find an inverse, just swap around the x & y variables then rearrange the equation:
^3})
^3} = 5 - x)
^3 = \dfrac{2}{5 - x})
^{1/3})
^{1/3})
Now for this equation to not have a divide-by-zero, you will need the following domain:
Domain:
(which is the range of our original function)
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