Question 687566
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There are three possibities with the coefficient of x<sup>4</sup> = 1

1. x=-4 has multiplicity 1 and x=-1 has multiplicity 3

f(x) = (x+4)(x+1)³
f(x) = (x+4)(x+1)(x+1)(x+1) 
       Multiply that out and get:
f(x) = x<sup>4</sup> + 7x<sup>3</sup> + 15x<sup>2</sup> + 13x + 4 


2. x=-4 has multiplicity 2 and x=-1 has multiplicity 2

f(x) = (x+4)²(x+1)²
f(x) = (x+4)(x+4)(x+1)(x+1) 
       Multiply that out and get:
f(x) = x<sup>4</sup> + 10x<sup>3</sup> + 33x<sup>2</sup> + 40x + 16 


3. x=-4 has multiplicity 3 and x=-1 has multiplicity 1

f(x) = (x+4)³(x+1)
f(x) = (x+4)(x+4)(x+4)(x+1) 
       Multiply that out and get:
f(x) = x<sup>4</sup> + 13x<sup>3</sup> + 60x<sup>2</sup> + 112x + 64 

Edwin</pre>