## anonymous one year ago Find a degree 3 polynomial with zeros -2, 1, and 5 going through the point (0,-3)

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1. anonymous

As a general rule when creating polynomials, if you are given ALL the zeros (i.e. the set of all x such that f(x)=0 ) and they are all real numbers, then you can form the polynomial by: $f(x)=\left( x + 2 \right)\left( x -1 \right)\left( x - 5 \right)$ Note if you plug in each of the zero's for the function one of the associated factors becomes zero, hence the whole function becomes zero. Also, note that you must know ALL the zeros for this to work, since complex zeros are possible for polynomials. Finally, while you can employ the same trick if you also know ALL the complex zeros (or at least one of every conjugate pair) you can do something similar; however I will omit this example since the question doesnt call for it. Now the remaining step in the problem is to force it to go through the point (0,-3). This can be done by first modifying f(x) as follows: $f(x)=c\left( x + 2 \right)\left( x -1 \right)\left( x - 5 \right)$ Where c is a constant to be determined. Now substitute in the point: $f(0)=-3=c\left( 2 \right)\left(-1 \right)\left(- 5 \right)=10c \ \ \ \rightarrow c=\frac{-3}{10}$ Therefore: $f(x)=\frac{-3}{10}\left( x + 2 \right)\left( x -1 \right)\left( x - 5 \right)$ Note none of the zeros of f(x) have changed since each factor still produces a zero upon substitution.