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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)

User Gnijuohz
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5.2k points

2 Answers

5 votes

Since the polynomial has roots of multiplicity 2 at x=5 and x=0, factors of (x−5) and( x=0 ) factors of
(x-5) ^2must be present. Also, since it has a root of multiplicity 1 at x=−1, the factor (x+1) must be present.

Since P(x) has a root of multiplicity 2 at x = 5, the factor
(x - 5)^2 must be present in the polynomial. Similarly, since P(x) has a root of multiplicity 2 at x = 0, the factor
x^2 must be present. Additionally, since P(x) has a root of multiplicity 1 at x = -1, the factor (x + 1) must be present.

Therefore, a possible formula for P(x) is:


P(x) = k(x - 5)^2 x^2 (x + 1)

where k is a constant. Since the leading coefficient of P(x) is 1, we can solve for k by substituting x = 0 and setting P(x) = 1:


1 = k(0 - 5)^2 0^2 (0 + 1)

Solving for k, we find k = 1.

Therefore, a possible formula for P(x) is:


P(x) = (x - 5)^2 x^2 (x + 1)

Complete question below:

The polynomial of degree 5, 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).

User Oleksandr Yefymov
by
5.3k points
3 votes

Answer:


P(x) = x^5 -9x^4 + 10x^3 + 25x^2

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

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 ,
(x-a)^m is the factor of the polynomial.


(x-5)^2
is a factor of P(x)


(x-0)^2
is another 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)

Solving the above expression , we get


P(x) = (x^2 + 25 -10x) (x^3 + x^2)  = x^3(x^2 + 25 -10x) +x^2(x^2 + 25 -10x) \\= x^5 + 25 x^3 -10x^4 + x^4 +25x^2 -10x^3 \\=x^5 -9x^4 + 10x^3 + 25x^2

Hence,
P(x) = x^5 -9x^4 + 10x^3 + 25x^2

User Lars Fosdal
by
5.2k points