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?

reyna
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mtom
  • 43

2 Answers2

3

$6(x-a)(x-b)(x-c)(x-d)(x-e) = 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$

Plugging x = -1

$6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e) = -6(a+1)(b+1)(c+1)(d+1)(e+1) = -6 + 5 - 4 + 3 - 2 + 1 = -3\\ (a+1)(b+1)(c+1)(d+1)(e+1) = \frac 12$

Doug M
  • 57,877
1

It is just $$-{1\over 6}p(-1)= -(-6+5-4+3-2+1) = {1\over 2}$$

Proof: $$p(-1) = 6(-1-a)(-1-b)(-1-c)(-1-d)(-1-e)$$ $$ = -6(1+a)(1+b)(1+c)(1+d)(1+c)$$

nonuser
  • 90,026