The polynomial of degree 4, P ( x ) P ( x ) has a root of multiplicity 2 at x = 2 x = 2 and roots of multiplicity 1 at x = 0 and x = − 2. It goes through the point ( 5 , 126 )

Answer 1

P(x)= a(x-2)^2(x)(x+2)

plug in point to find "a"

126 = a(3)^2)(5)(7)

a= 126/9(35)= 14/35=2/5

P(x) = (2/5)(x^2 -4x+4)(x^2 +2x)

= .4(x^4 -2x^3 -4x^2 +8x)

= .4x^4 -.8x^3 -1.6x^2 +3.2x

Answer 2

Saying that P(x) is a degree 4 polynomial. means its highest term will be ax⁴.

Saying it has a root of multiplicity 2 at x = 2 means that it will have a two factors of (x - 2)

Saying it has a root of multiplicity 1 at x = 0 and x = -2 means it will have factors at (x - 0) and (x - -2) or, simplified (x) and (x + 2).

Since we have listed 4 factors, each having a "x" in it, we will achieve our x⁴ requirement which will give us:

P(x) = (x - 2)(x - 2)(x)(x + 2) however, that does NOT meet our ax⁴ requirement so we MUST include a multiplier "a" in front of our polynomial:

P(x) = a(x - 2)(x - 2)(x)(x + 2)

To determine the value of "a", we must use the point given to us (5, 126). This means that when x = 5, P(x) will equal 126 so:

126 = a(5 - 2)(5 - 2)(5)(5 + 2)

126 = a(3)(3)(5)(7)

126 = 315a

a = 126/315 = 2/5

So P(x) = 2/5(x - 2)(x - 2)(x)(x + 2)

If you'd like, you can multiply this out but I would not think it would be required.

Answer 3

P(x) = a(x - 2)²x(x + 2); P(5) = a(5 - 2)²·5·(5 + 2) = 126; a = 126/(63·5 = 2/5 = 0.4;

P(x) = 0.4(x - 2)²x(x + 2)