Find a polynomial p(x) with real coefficients having the given zeros.
6,-14, and 4+3i
4-3i must also be a root, so we have
p(x) = (p-6)(p+14)(p-(4+3i))(p-(4-3i))
= (p-6)(p+14)((p-4)^2 + 3^2)
= (p-6)(p+14)(p^2-8p+25)
= p^4 - 123p^3 + 872p - 2100
Perfect Thanks,
To find a polynomial with the given zeros, we can use the fact that if a number "a" is a zero of the polynomial, then (x - a) will be a factor of that polynomial.
Given zeros: 6, -14, and 4+3i
Since 4+3i is a complex number, its conjugate 4-3i will also be a zero. Hence, the zeros are: 6, -14, 4+3i, and 4-3i.
To build the polynomial, we can write it as a product of its linear factors:
p(x) = (x - 6)(x + 14)(x - (4+3i))(x - (4-3i))
To simplify further, let's expand and multiply the factors:
p(x) = (x - 6)(x + 14)((x - 4) - 3i)((x - 4) + 3i)
Using the difference of squares pattern:
p(x) = (x - 6)(x + 14)((x - 4)^2 - (3i)^2)
Expanding further:
p(x) = (x - 6)(x + 14)((x - 4)^2 - 9i^2)
Since i^2 is equal to -1:
p(x) = (x - 6)(x + 14)((x - 4)^2 - 9(-1))
Simplifying the expression:
p(x) = (x - 6)(x + 14)(x^2 - 8x + 16 + 9)
Finally, let's distribute and simplify the expression:
p(x) = (x - 6)(x + 14)(x^2 - 8x + 25)
Therefore, the polynomial p(x) with real coefficients having the given zeros 6, -14, and 4+3i is:
p(x) = (x - 6)(x + 14)(x^2 - 8x + 25)