0
Since the polynomial has three zeroes it has a degree of 3. So we have:
(x - -5)[x - (2 + i)][x - (2 - i)]
= (x + 5)[(x - 2) - i][(x - 2) + i]
= (x + 5)[(x - 2)2 - i2]
= (x + 5)[x2 - 2(x)(2) + 22 - (-1)]
= (x + 5)(x2 - 4x + 4 + 1)
= (x + 5)(x2 - 4x + 5)
= x(x2 - 4x + 5) + 5(x2 - 4x + 5)
= x3 - 4x2 + 5x + 5x2 - 20x + 25
= x3 + x2 - 15x + 25
Thus,
P(x) = x3 + x2 - 15x + 25
What else can I help you with?
What does 7i divided by 2 plus 2i equal?
1.75 + 1.75i7i/2 + 2i = 3.5i + 2i = 5.5i
What is the answer if you Find all the roots of the equation x4 - 2x3 plus 14x2 - 18x plus 45 0 given that 1 plus 2i is one of its roots.?
The four roots are:1 + 2i, 1 - 2i, 3i and -3i.
Which is equal to -5 plus 6i - 7-8i?
-5-7=-126i-8i=-2i-12-2i is your final answer
What is the magnitude of the complex number 8 plus 2i?
8.2462 (rounded)
What is 2-4i over 4 plus 2i put in the form of a plus bi?
To solve this type of problem, multiply both the numerator and denominator by the conjugate of the denominator. (2 - 4i) / (4 + 2i) = (2 - 4i)(4 - 2i) / (4 + 2i)(4 - 2i) then expand all the terms, and simplify. = (8 - 20i + 8i2) / (16 - 4i2) = (8 - 20i - 8) / (16 + 4) = -20i / 20 = -i Which in the required answer format becomes, 0 + i.
What is the quartic polynomial function with rational coefficients that has roots you and 2i?
(x - u)*(x - u)*(x + 2i)*(x - 2i) = (x2 - 2xu + u2)*(x2 + 4) = x4 - 2x3u + x2(u2 + 4) - 8xu + 4u2
If 3 - 2i is a solution to a polynomial equation is the complex conjugate 3 plus 2i also a solution yes or no?
Not necessarily, take for example the equation x^2=5-12i. Then, 3-2i satisfies the equation. However, 3+2i does not because (3+2i)^2 = 5+12i.
What are the zeros of this polynomial function f of x equals x3 minus 3x2 minus 5x plus 39?
f(x)=x3-3x2-5x+39=(x+3)(x2-6x+13) It has three roots. One of which is x=-3. Using the quadratic equation: x = (6 +/- √(-16))/2 x = (6 +/- 4i)/2 = (3 +/- 2i) so, x=-3, x=3+2i, or x=3-2i
What is the equivalent form of 10 plus 6i and 7 plus 2i?
10 + 6i and 7 + 2i = 10 + 6i + 7 + 2i = 17 + 8i
What does 7i divided by 2 plus 2i equal?
1.75 + 1.75i7i/2 + 2i = 3.5i + 2i = 5.5i
Is it true that the degree of polynomial function determine the number of real roots?
Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)
The conjugate of the complex number 3 plus 2i is?
It is 3 minus 2i
What is a polynomial with a degree of 3 and zeros of -3 and 3-2i?
The question has a simple answer only if the polynomial has rational coefficients. However, the question does not state that it has rational coefficients so it is not valid to assume that is the case. So, suppose the third root is p. Then the polynomial is (x + 3)*(x - 3 + 2i)*(x - p) = (x2 + 2xi - 9 + 6i)*(x - p) = x3 + (2i - p)x2 - (2pi - 6i + 9)x + (9p + 6pi)
What is the answer if you Find all the roots of the equation x4 - 2x3 plus 14x2 - 18x plus 45 0 given that 1 plus 2i is one of its roots.?
The four roots are:1 + 2i, 1 - 2i, 3i and -3i.
Find the sum of negative 2 plus 3i and negative1 minus 2i?
(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i
Which is equal to -5 plus 6i - 7-8i?
-5-7=-126i-8i=-2i-12-2i is your final answer
What is the complete ionic equation for h2so4 plus cai2 equals caso4 plus 2hi?
The complete ionic equation for this reaction is: 2H⁺(aq) + SO₄²⁻(aq) + Ca²⁺(aq) + 2I⁻(aq) → CaSO₄(s) + 2H⁺(aq) + 2I⁻(aq)
