Question

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)

Answer 1

Answer:

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

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$x^5 + bx^4 + cx^3 + dx^2 + ex + f$

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,  $P(x) = x^5 -9x^4 + 10x^3 + 25x^2$