We must find the zeros of the following function:
[tex]f(x)=x^3-x^2-11x+15.[/tex]
1) First, we plot a graph of the function:
From the graph, we see that the function crosses the x-axis at x = 3, so x = 3 is one of the zeros of the function.
2) Because x = 3 is a zero of the function, we can factorize the function in the following way:
[tex]f(x)=x^3-x^2-11x+15=(x^2+b\cdot x+c)\cdot(x-3)\text{.}[/tex]
To find the coefficients b and c, we compute the product of the parenthesis and then we compare the different terms:
[tex]f(x)=x^3-x^2-11x+15=x^3+(b-3)\cdot x^2+(c-3b)\cdot x-3c.[/tex]
To have the same expressions at both sides of the equality we must have:
[tex]\begin{gathered} -3c=15\Rightarrow c=-\frac{15}{3}=-5, \\ b-3=-1\Rightarrow b=3-1=2. \end{gathered}[/tex]
So we have the following factorization for the function f(x):
[tex]f(x)=(x^2+2x-5)\cdot(x-3)\text{.}[/tex]
3) To find the remaining zeros, we compute the zeros of:
[tex](x^2+2x-5)\text{.}[/tex]
The zeros of this 2nd order polynomial are given by:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2a},[/tex]
where a, b and c are the coefficients of the polynomial. In this case we have a = 1, b = 2 and c = -5. Replacing these values in the formula above, we get:
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