Question

can u please correct this problem for me i wanna to make sure that is right or wrong??? use the given zero to find the remaining zeros of the function: f(x)= x^5-10x^4+42x^3-124x^2+297x-306 ; zero: 3i.... my answer is -2276

Answer 1

I don't really understand your answer. You should find the other four numbers that make the polynomial zero (because it's a polynomial of degree 5, by the fundamental theorem of algebra it has exactly five zeroes).
Because we already know one zero, we can use long division to get a polynomial of degree four. -3i must also be a zero (because it's a polynomial with real coefficients is a complex number is a zero it's conjugate must be a zero too) so we can use long division again to find a polynomial of degree 3:
\[x^3 -10x^2 +33x - 34 = 0\]This polynomial must at least one real solution. If this solution is rational, it must be an integer (because it's a monic polynomial with integer coefficients) and must divide 34, so the options are\[\pm 1, \pm 2, \pm 17, \pm 34.\]By testing these numbers we find that 2 is a zero, then we use long division yet again to find\[x^2-8x+17\]and we use the quadratic formula to find the two remaining zeros: 4 + i and 4 -i. Therefore, the zeros of the function are\[2, 3i, -3i, 4 -i, 4 + i.\]

Answer 2

use the given zero to find the remaining zeros of the function : f(X)= x^5-10x^4+42x^3-124x^2+297x-306..zero:3i

Answer 3

@Krebante

Can u pls calrify with what did you divide the original polynomial to get the cubic quotient polynomial??

Pls explain your first step in detail as I am interested in learning this....☺
Thanks in advance...☻

Answer 4

Because 3i is a root of the polynomial, and all the coefficients of the polynomial are real numbers, then -3i must also be a root. So basically we know that the product:
\[(x-3i)(x+3i) = x^{2}+9\]
is a factor of the polynomial. To get that cubic polynomial, you would divide that 5th degree polynomial by x^2+9

Answer 5

\[f(x) = (x-2) \left(x^2+9\right) \left(x^2-8 x+17\right) \]

Answer 6

long division is gross >.>

Answer 7

Let \[p(z) = \sum_{k = 0}^n a_n z^k\]a polynomial with complex coefficients. Because of the fundamental theorem of algebra, it has exactly n zeros (some of which may have multiplicity greater than 1). Let\[z_1\]such that\[p(z_1) = 0.\]Because the polynomials are an euclidean domain, we can find two polynomials\[q_1(z)\]and\[r_1(z)\]such that\[p(z) = q(z)(z - z_1) + r(z).\]By the polynomial remainder theorem,\[r_1(z) = p(z) = 0,\]so\[p(z) = q_1(z)(z-z_1)\]and we can divide p(z) by\[(z-z_1)\]with remainder zero:\[q_1(z) = \frac{p(z)}{z-z_1}\]and the degree of\[q_1(z)\]must be exactly n - 1, so it has n - 1 zeros. We repeat this process with every root and finally we have\[p(z) = a_n(z-z_1)^{m_1} \dotsm (z - z_\ell)^{m_\ell}\]where\[\sum_{k = 1}^\ell m_k = n.\]Which means the polynomial can be factored as a multiplication of polynomials of degree 1. This is exactly what I did: because 3i is a root we can divide the polynomial by x - 3i and find a polynomial of degree 4. Then, -3i must also be a root (because if a complex number is a zero a polynomial with real coefficients it's conjugate must be a zero too) so we can divide by x + 3i and find a polynomial of degree 3. We can find the remaining zeroes of the original polynomial by finding the zeroes of this polynomial of degree 3. If you want to know the exact process for dividing a polynomial by a polynomial of the form x - a, I suggest Ruffini's algorithm: http://en.wikipedia.org/wiki/Ruffini's_rule#Polynomial_division_by_x_.E2.88.92_r

Answer 8

Thanks everybody for the various explanations........
I understand it now.....☺☻☺