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Verdich [7]
3 years ago
11

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

multiplicity 1 at x = − 1 Find a possible formula for P ( x ) .
Mathematics
1 answer:
ser-zykov [4K]3 years ago
5 0

Answer:

p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}, for r_{1} = 0

Step-by-step explanation:

The general form of quintic-order polynomial is:

p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f

According to the statement of the problem, the polynomial has the following roots:

p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)

Then, some algebraic handling is done to expand the polynomial:

p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)

p_{5} (t) = x^{5} - (5+r_{1})\cdot x^{4} + (3 + 5\cdot r_{1})\cdot x^{3} +(9-3\cdot r_{1})\cdot x^{2} - 9 \cdot r_{1}\cdot x

If r_{1} = 0, then:

p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}

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