Polynomial equation

Question

The polynomial of degree 4, P(x)P(x) has a root of multiplicity 2 at x=1x=1 and roots of multiplicity 1 at x=0x=0 and x=−1x=-1. It goes through the point (5,240)(5,240). 

Find a formula for P(x)P(x).

Answer 1

EXPLANATION:

From given conditions we make polynomial of degree 4.

P(x)= x(x-1)²(x+1)

=x(x²-2x+1)(x+1)

=x(x²-2x+1).x+(x²-2x+1).1

=x(x³ -2x²+x +x² -2x +1)

=x(x³-x²-x+1)

P(x)=x⁴ -x³ -x² +x

We need to fix it up further to get it through the given point (5, 240)

P(x)= b(x⁴-x³-x²+x)...........(A)

240= P(5) = m[(5)⁴ - (5)³ - (5)² + 5]

240 = m(625 -125 -25+5)

240 = m(480)=

m= 240/480 = 1/2

Put value of m in (A)

P(x) = (1/2)(x⁴ -x³ -x³+x)

P(x) = x⁴/2 - x³/2 -x²/2 + x/2

Answer 2

x(x-1)^2(x+1)

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

=x(x^3 -x^2 -x +1)

= x^4 -x^3 -x^2 +x

which has

zeros 0, 1 and -1

1 has multiplicity 2, indicated by the exponent for the corresponding factor

but you need to fix it up further to get it through the given point (5, 240)

240 = P(5) = a[(5)^4 - (5)^3 - (5)^2 + 5]

solve for a

240 = a(625 -125 -25+5)

240 = a(480)

a = 240/480 = 1/2

P(x) = (1/2)(x^4 -x^3 -x^2 +x)

P(x) = x^4/2 - x^3/2 -x^2/2 + x/2