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?
If the other root is k,
f(x) = (x+5)(x+3)(x-4)(x-k)
= x^4 + (4-k)x^3 - (4k+17)x^2 + (17k-60)x + 60k
So,
4-k = -2
k=6
a = -(4k+17) = -41
b = 17k-60 = 42
c = 60*6 = 360
a+b+c = 361
thanks STEVE...
To find the value of a+b+c, we need to determine the coefficients a, b, and c of the polynomial function f(x)=x^4−2x^3+ax^2+bx+c.
We know that the roots of the polynomial are -5, -3, and 4.
Since the roots are given, we can set up equations using Vieta's formulas. Vieta's formulas state that for a polynomial function with roots r1, r2, r3, and r4, the following relationships hold:
r1 + r2 + r3 + r4 = -b/a
r1 * r2 * r3 * r4 = c/a
In this case, we have three roots: -5, -3, and 4. So we can write the following equations:
-5 + (-3) + 4 + r4 = -(-2)/1
-5 * (-3) * 4 * r4 = c/1
Simplifying the equations, we get:
-4 + r4 = 2
60r4 = c
From the first equation, we can solve for r4:
-4 + r4 = 2
r4 = 2 + 4
r4 = 6
Now, substituting this value of r4 into the second equation:
60 * 6 = c
c = 360
Therefore, the value of c is 360.
Since we know that a+b+c is the sum of the coefficients of the polynomial, we have:
a+b+c = a + b + 360
Since we don't have enough information to determine the values of a and b, we cannot solve for a+b+c without additional information.