f(x)is a monic cubic polynomial with real coefficients. f(x) has 7+4i as a root and f(0)=−2145 . What is the coefficient of x in f(x) ?

as you know, the complex roots come in conjugate pairs. So, the other root is 7-4i

f(x) = (x-(7+4i))(x-(7-4i))(x-k)
= (x^2-14x+65)(x-k)

f(0) = 65(-k) = -2145
so, k = 33

f(x) = (x^2-14x+65)(x-33)
= x^3 - 47x^2 + 6527x - 2145

you mean 527.

Thanks a lot

Dang. A typo in the only place it mattered!

To find the coefficient of x in the monic cubic polynomial f(x), we need to find the other two roots of the polynomial.

Given that f(x) has a root of 7+4i, we know that its complex conjugate, 7-4i, must also be a root. This is because complex roots always come in conjugate pairs for polynomials with real coefficients.

To find the third root, we can use the fact that the sum of the roots of a cubic polynomial is equal to the opposite of the coefficient of the quadratic term (x^2) divided by the coefficient of the cubic term (1 in this case).

Since f(x) is monic (meaning its leading coefficient is 1), we can write the polynomial as:
f(x) = (x - (root1))(x - (root2))(x - (root3))

We already know root1 = 7+4i and root2 = 7-4i. Let's say root3 = a (a real number).

Substituting these values into the polynomial:

f(x) = (x - (7+4i))(x - (7-4i))(x - a)

Now, we can use the fact that f(0) = -2145 to solve for a:

f(0) = (0 - (7+4i))(0 - (7-4i))(0 - a)
-2145 = (7+4i)(7-4i)(-a)

Expanding the equation:

-2145 = (49 - (4i)^2)(-a)
-2145 = (49 - 16i^2)(-a)
-2145 = (49 - 16(-1))(-a)
-2145 = (49 + 16)(-a)
-2145 = 65(-a)
-2145 = -65a

Now, we can solve for a:

-2145 / -65 = a
33 = a

So, the third root is a = 33.

Now, to find the coefficient of x in f(x), we can multiply out the polynomial:

f(x) = (x - (7+4i))(x - (7-4i))(x - 33)
f(x) = (x^2 - (7+4i)x - (7-4i)x + (7+4i)(7-4i))(x - 33)
f(x) = (x^2 - 14x + 65)(x - 33)
f(x) = x^3 - 47x^2 + 587x - 2145

As a result, the coefficient of x in f(x) is -47.