The given equation is a quartic equation, and a, b, r, s are its roots. According to Vieta's formulas, the sum of the roots taken one at a time is -b/a, the sum of the roots taken two at a time is c/a, the sum of the roots taken three at a time is -d/a, and the product of the roots is -e/a.

In this case, the equation is x^4 + x^3 + x^2 + x + 3 = 0. So, a = 1, b = 1, c = 1, d = 1, and e = 3.

The expression (1+a)(1+b)(1+r)(1+s) is the product of the roots of the equation increased by 1.

We can rewrite this expression as (1 - (-a))(1 - (-b))(1 - (-r))(1 - (-s)), which is the product of the roots of the equation x^4 - x^3 + x^2 - x + 1 = 0, increased by 1.

According to Vieta's formulas, the product of the roots of this equation is -e/a = -1/1 = -1.

So, the value of (1+a)(1+b)(1+r)(1+s) is -1.

Question

The given equation is a quartic equation, and a, b, r, s are its roots. According to Vieta's formulas, the sum of the roots taken one at a time is -b/a, the sum of the roots taken two at a time is c/a, the sum of the roots taken three at a time is -d/a, and the product of the roots is -e/a.

In this case, the equation is x^4 + x^3 + x^2 + x + 3 = 0. So, a = 1, b = 1, c = 1, d = 1, and e = 3.

The expression (1+a)(1+b)(1+r)(1+s) is the product of the roots of the equation increased by 1.

We can rewrite this expression as (1 - (-a))(1 - (-b))(1 - (-r))(1 - (-s)), which is the product of the roots of the equation x^4 - x^3 + x^2 - x + 1 = 0, increased by 1.

According to Vieta's formulas, the product of the roots of this equation is -e/a = -1/1 = -1.

So, the value of (1+a)(1+b)(1+r)(1+s) is -1.

Knowee AI · Accepted Answer

The given equation is a quartic equation, and a, b, r, s are its roots. According to Vieta's formulas, the sum of the roots taken one at a time is -b/a, the sum of the roots taken two at a time is c/a, the sum of the roots taken three at a time is -d/a, and the product of the roots is -e/a.

In this case, the equation is x^4 + x^3 + x^2 + x + 3 = 0. So, a = 1, b = 1, c = 1, d = 1, and e = 3.

The expression (1+a)(1+b)(1+r)(1+s) is the product of the roots of the equation increased by 1.

We can rewrite this expression as (1 - (-a))(1 - (-b))(1 - (-r))(1 - (-s)), which is the product of the roots of the equation x^4 - x^3 + x^2 - x + 1 = 0, increased by 1.

According to Vieta's formulas, the product of the roots of this equation is -e/a = -1/1 = -1.

So, the value of (1+a)(1+b)(1+r)(1+s) is -1.

If a, b, r, s are roots of x^4+x^3+x^2+x+3=0 Find value of (1+a)(1+b)(1+r)(1+s)

Question

If a, b, r, s are roots of $x^4 + x^3 + x^2 + x + 3 = 0$ Find value of $(1+a)(1+b)(1+r)(1+s)$

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