Write a polynomial of the lowest degree that for x = 0 takes the value of 1 and has the following roots: 1 and -1 as simple roots, 2 as a double root, -3 as a triple root.
Let [tex] p(x) [/tex] be the polynomial. It is given that it has the following roots/zeroes: 1 and -1 as simple roots, 2 as a double root, -3 as a triple root. Note that if [tex]p(x)[/tex] has a zero [tex] \alpha[/tex], then it has a factor [tex] (x - \alpha)[/tex] and vice versa. For the given problem, we can write [tex] p(x) [/tex]:
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Verified answer
ANSWER:
[tex] \boxed{p(x) = -\dfrac{1}{108}(x - 1)(x + 1)(x - 2)^2(x + 3)^3} [/tex]
SOLUTION:
Let [tex] p(x) [/tex] be the polynomial. It is given that it has the following roots/zeroes: 1 and -1 as simple roots, 2 as a double root, -3 as a triple root. Note that if [tex]p(x)[/tex] has a zero [tex] \alpha[/tex], then it has a factor [tex] (x - \alpha)[/tex] and vice versa. For the given problem, we can write [tex] p(x) [/tex]:
[tex] p(x) = k(x - 1)(x + 1)(x - 2)^2(x + 3)^3 [/tex], where [tex] p(0) = 1.[/tex]
Solving for [tex] k,[/tex]
[tex] \begin{array}{c} P(0) = k(-1)(1)(-2)^2(3)^3 = 1 \\ -108k = 1 \\ k = -\dfrac{1}{108} \end{array} [/tex]
Thus,
[tex]\boxed{p(x) = -\dfrac{1}{108}(x - 1)(x + 1)(x - 2)^2(x + 3)^3} [/tex]