f(x)=x^4+ax^3+bx^2+cx+d and has real coefficients. If two of the roots are 4−i and 2+3i, what is the value of a+b+c+d?
complex roots always come in conjugate pairs. So,
f(x) = (x-(4-i))(x-(4+i))(x-(2+3i))(x-(2-3i))
= ((x-4)^2 + 1^2)((x-2)^2 + 3^2)
= (x^2-8x+17)(x^2-4x+13)
= x^4 - 12x^3 + 62x^2 - 172x + 221
...
To find the value of the coefficients a, b, c, and d, we can use the fact that complex roots occur in conjugate pairs.
Given that the roots are 4 - i and 2 + 3i, their conjugates are 4 + i and 2 - 3i respectively.
Now, the polynomial can be written as:
f(x) = (x - (4 - i))(x - (4 + i))(x - (2 + 3i))(x - (2 - 3i))
Multiplying out the factors, we have:
f(x) = (x - 4 + i)(x - 4 - i)(x - 2 - 3i)(x - 2 + 3i)
Expanding each pair of conjugates using the difference of squares formula, we get:
f(x) = ((x - 4)^2 - i^2)((x - 2)^2 - (3i)^2)
Applying the simplification, we have:
f(x) = (x^2 - 8x + 16 + 1)(x^2 - 4x + 4 + 9)
Simplifying further, we get:
f(x) = (x^2 - 8x + 17)(x^2 - 4x + 13)
Expanding this, we get:
f(x) = x^4 - 12x^3 + 61x^2 - 116x + 221
Comparing this polynomial to the given polynomial f(x) = x^4 + ax^3 + bx^2 + cx + d, we can equate the corresponding coefficients:
a = -12
b = 61
c = -116
d = 221
Therefore, the value of a + b + c + d is:
a + b + c + d = -12 + 61 - 116 + 221 = 154