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Madeline6117
27.11.2019 •
Mathematics
The polynomial of degree 5, p(x) p(x) has leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=−1
find a possible formula for p(x)
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Ответ:
Step-by-step explanation:
The given degree of the polynomial P(x) = 5
The leading coefficient = 1
So, the general form of the polynomial with degree 5 is![x^5 + bx^4 + cx^3 + dx^2 + ex + f](/tpl/images/0392/9138/05926.png)
Now root x =5 is of multiplicity 2, x = 0 of multiplicity 2, x = -1 of multiplicity 1
If x = a is the zero of the polynomial of multiplicity m, then ,
is the factor of the polynomial.
⇒
is a factor of P(x)
(x +1) is the last factor of P(x)
So, P(x) = Product of all factors =![(x-5)^2 (x)^2(x+1)](/tpl/images/0392/9138/8f7eb.png)
Solving the above expression , we get
Hence,![P(x) = x^5 -9x^4 + 10x^3 + 25x^2](/tpl/images/0392/9138/4de76.png)
Ответ: