Three roots of f(x)=x^4-2x^3 +ax^2 +bx+c are −5 , −3 and 4 . What is the value of a+b+c ?
To find the value of a+b+c, we need to use the concept of Vieta's formulas. Vieta's formulas state that for a polynomial function f(x) of degree n, the sum of the roots is equal to the negation of the coefficient of the second-to-last term divided by the coefficient of the leading term.
In this case, the polynomial f(x) has degree 4, so the sum of the roots is given by:
Sum of roots = - (coefficient of second-to-last term) / (coefficient of leading term)
The coefficient of the second-to-last term is the coefficient of the x^2 term, which is 'a'. The coefficient of the leading term is the coefficient of the x^4 term, which is 1.
Therefore, the sum of the roots is -a/1 = -a.
Since we know that the roots are -5, -3, and 4, we can substitute these values into the equation:
-5 - 3 + 4 = -a
Simplifying, we get:
-4 = -a
To find the value of 'a', we can multiply both sides of the equation by -1:
4 = a
Now, we have found the value of 'a'.
To find the value of 'b', we can use another Vieta's formula which states that the product of the roots taken two at a time is equal to the constant term divided by the leading coefficient.
In this case, the product of the roots taken two at a time is equal to:
Product of roots taken two at a time = constant term / leading coefficient
The constant term is 'c' and the leading coefficient is 1.
Therefore, the product of the roots taken two at a time is equal to c/1 = c.
Since we know that the roots are -5, -3, and 4, we can substitute these values into the equation:
(-5 * -3) + (-5 * 4) + (-3 * 4) = c
Simplifying, we get:
15 + (-20) + (-12) = c
3 = c
Now, we have found the value of 'c'.
Finally, to find the value of a+b+c, we can substitute the values of 'a', 'b', and 'c' into the expression:
a + b + c = 4 + b + 3
Simplifying, we get:
a + b + c = 7 + b
Since 'b' is not given, we do not have enough information to find its value. Therefore, the value of a+b+c cannot be determined with the given information.