Question

If a, b, c, d, e are all zeroes of the polynomial (6x
^
5 + 5x
^
4 + 4x
^
3 + 3x
^
2 + 2x + 1), find

the value of (1 + a) (1 + b) (1 + c) (1 + d) (1 + e)​

Answer 1

Answer:

1/2

Step-by-step explanation:

Expanding (1+a)(1+b)(1+c)(1+d)(1+e) we get formula:

1+(a+b+c+d+e) +(ab+ac+ad+ae+bc+bd+be+cd+ce+de) + (abc+abd+abe+ace+ade+acd+bcd+bce+bde+cde) +(abcd+abce+bcde+cdae+abde) + abcde    ------- (1)

in terms of:

6$x^5$ + 5$x^4$ + 4x³ + 3x² + 2x + 1 =0

$x^5$ + (5/6)$x^4$ + (4/6)x³ + (3/6)x² + (2/6)x + (1/6)x = 0

$x^5$ - (Sum of roots)$x^4$ + (Sum of roots taken 2 at a time)x³ - (Sum of roots taken 3 at a time)x² + (Sum of roots taken 4 at a time)x - (Product of all 5 roots) = 0    ------- (2)

Now, comparing like terms we get the value for (1+a)(1+b)(1+c)(1+d)(1+e) as:

1 - (5/6) + (4/6) - (3/6) + (2/6) - (1/6)

= (1/2)