<s>[INST] $a,b,c,d,e$ are zeroes of $6x^5+5x^4+4x^3+3x^2+2x+1$ find $(a+1)(b+1)(c+1)(d+1)(e+1)$ 
If $a,b,c,d,e$ are zeroes of the polynomial
  $$6x^5+5x^4+4x^3+3x^2+2x+1$$
  find the value of $(a+1)(b+1)(c+1)(d+1)(e+1)$.

According to me in order to solve this problem one should first factorize the given polynomial in the form of:
$$(x-a)(x-b)(x-c)(x-d)(x-e)$$
and equate the two and then compare the coefficients to get the following equations:
$(a+b+c+...)=-5$
$(ab+ac+ad+bd+.....)=4$
$(abc+abd+abe+...)=-3$
$(abcd+abce+abde+.....)=2$
$abcde=-1$
Now, adding all the equations and with some factorization we get:
$$(a+1)(b+1)(c+1)(d+1)(e+1)=-2$$
but my book says the answer is $0.5$.
Where am I going wrong and what is the actual method to solve this, with the correct answer?
 [/INST] It is just $$-{1\over 6}p(-1)= -(-6+5-4+3-2+1) = {1\over 2}$$
Proof: $$p(-1) = 6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e)$$
$$ = -6(1+a)(1+b)(1+c)(1+d)(1+c)$$
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