# Given the polynomial f(x)=x^6+4x^5+x^4-12x^3-11x^2+4x+4 has a zero of multiplicity 2 at x=-2, use the reminder theorem to find out the quotient.

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### 1 Answer

We'll recall the reminder theorem:

f(x) = (x+2)^2*Q(x) + 0

Since x = -2 is the root of the polynomial, then the reminder is 0.

Since the order of multiplicity of the root is 2, then the quotient is a polynomial of degree 4.

f(x) = (x+2)^2*(ax^4+bx^3+cx^2+dx+e)

To determine the quotient, we'll have to calculate the coefficients a,b,c,d,e.

We'll expand the square:

f(x) = (x+2)^2*(ax^4+bx^3+cx^2+dx+e)

f(x) = (x^2 + 4x + 4)*(ax^4+bx^3+cx^2+dx+e)

We'll remove the brackets:

f(x) = ax^6 + bx^5 + cx^4 + dx^3 + ex^2 + 4ax^5 + 4bx^4 + 4cx^3 + 4dx^2 + 4ex + 4ax^4 + 4bx^3 + 4cx^2 + 4dx + 4e

f(x) = ax^6 + x^5*(4a+b) + x^4*(c + 4b + 4a) + x^3*(d + 4c + 4b) + x^2*(e + 4d + 4c) + x*(4e + 4d) + 4e

Comparing both sides, we'll get:

a =1

4a + b = 4 => b = 0

c + 4b + 4a = 1 => c + 4 = 1 => c = -3

d + 4c + 4b = -12 => d - 12 = -12 => d = 0

e + 4d + 4c = -11 => e - 12 = -11 => e = 1

4e + 4d = 4

e + d = 1 => e = 1

**The requested quotient is: Q(x) = x^4 - 3x^2 + 1.**