Given that the polynomial function`f(x)=6-x-x^2 `is a factor of the polynomial `g(x)=ax^3+5x^2+bx-18`, find the values of the constants a and b.

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The function `f(x)=6-x-x^2 ` can be factored as `f(x) = (3+x)(2-x)` . Since `f(x)` is a factor of `g(x)` , each of the factors of `f(x) ` , `3+x` and `2-x` , is also a factor of `g(x)` .

According to the Fundamental Theorem of Algebra, if expression `x-a` is a factor of the polynomial `P(x)` , then `P(a) = 0` .

This means that `g(x) = 0 ` for `x=2` and `x=-3` .

Plug in these values into the expression for `g(x)` :

`g(2)=a(2^3)+5*(2^2) +b*2-18 = 0`

`g(-3)=a(-3)^3+5*(-3)^2+b*(-3) - 18=0`

The above yields two equations for `a` and `b` :

`8a+2b=-2`

`-27a-3b=-27`

Dividing first equation by 2 and second by 3 yields

`4a+b=-1`

`-9a-b=-9`

Adding the two equations results in

`-5a=-10` , or `a = 2` .

From the first equation, `b=-1-4a=-9` .

**Therefore a = 2 and b = -9.**

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