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Given that the polynomial function`f(x)=6-x-x^2 `is a factor of the polynomial...
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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` :
Dividing first equation by 2 and second by 3 yields
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.
Posted by ishpiro on July 28, 2013 at 2:32 AM (Answer #1)
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