1)A.f(x)= x^2+4
B.f(x)= x^2-4x^2+4x-16
C.f(x)= x^2+4x^2+4x+16
D.f(x)= x^2-4x^2-4x+16
2)A.+-1,+-2,+-3,+-6
B.0,+-1,+-2,+-3,+-6,+-1/3,+-2/3
C.+-1,+-2,+-3,+-6,+-1/3,+-2/3
D.+-1,+-3,+-1/6,+-1/3,+-1/2,+-3/2
4)I don't know what they mean either but this is all it says.
A.3x^2-x+5
B.75-30x+3x^2
C.3x^2-15x^2
D.15x^2-3x^3
what do they mean when they say like 2i,what is that?
1)Write a polynomial function of least degree with integral coefficients whose zeros include 4 and 2i.
answer= f(x)= x^2-4x^2+4x-16
2)List all of the possible rational zeros of f(x)= 3x^3-2x^2+7x+6
dont know
3)Find all of the rational zeros of f(x)= 4x^3-3x^2-22x-15
dont know
4)Find (f.g)(x) for f(x)= 3x^2 and g(x)= 5-x
answer= 3x^2-15x^2
1. imaginary roots always come in pairs, like in ±2i
so the factors would be (x+2i)(x-2i)(x-4)
expand it and you will have your answer.
2. I tried the factor theorem hoping for some f(a)=0 where a=±1,±2,±3
None worked so I don't know how you are expected to do that one.
You could try ±1/3,±2/3 but that seems a bit too farfetched from the type of questions you seem to have
3. try f(-1) it will be a zero
so x+1 is a factor. Do synthetic division or long division, you should get an answer of 4x^2 - 7x - 15 which factors again.
(see if you can get zeros at x=-1,3,-5/4
4. I don't know if your textbook defines
(f∙g) as f(g(x)) or g(f(x)).
f(g(x)) = f(5-x) = 3(5-x)^2
= 75 - 30x + 3x^2
g(f(x)) = g(3x^2) = 5 - 3x^2
i is the symbol for √(-1), which is the imaginary unit number.
so 2i is really 2√(-1)
in other words i^2 = -1
eg. solve x^2 + 9=0
x^2 = -9
x = ±(√9)(√(-1))
x = ±3√(-1)
x = ±3i
1. After I expanded my answer to #1 above I got x^3 - 4x^2 + 4x - 16 which is the same as B if your first term is x^3, as it should be for all of those answers.
2. for this questions according to the answers I can see that they simply wanted those values that your would try in your f(x) function. As I noted in my first answer, the correct answer would appear to be C
4. So it is B
Look at my first solution how I got that.
1) The polynomial function of least degree with integral coefficients whose zeros include 4 and 2i is obtained by multiplying the factors (x - 4), (x + 2i), and (x - 2i) together. By simplifying the expression, we get f(x) = x^2 - 4x^2 + 4x - 16.
2) To find the possible rational zeros of f(x) = 3x^3 - 2x^2 + 7x + 6, we can use the Rational Root Theorem. According to the theorem, the possible rational zeros are of the form p/q, where p is a factor of the constant term (in this case, 6) and q is a factor of the leading coefficient (in this case, 3). By listing all the factors of 6 (±1, ±2, ±3, ±6) and 3 (±1, ±3), we can check which ones are zeros of the polynomial by substituting them into f(x) and solving for zero. Unfortunately, there is no simpler way to determine the exact rational zeros, so we have to try each possibility.
3) Similarly to question 2, we can use the Rational Root Theorem to find the rational zeros of f(x) = 4x^3 - 3x^2 - 22x - 15. The factors of 15 are ±1, ±3, ±5, and ±15, and the factors of 4 are ±1 and ±2. By substituting these potential rational zeros into f(x) and checking for zero, we can find the rational zeros of the polynomial.
4) The notation (f.g)(x) denotes the composition of two functions, f(x) and g(x). In this case, f(x) = 3x^2 and g(x) = 5 - x. To find (f.g)(x), we substitute g(x) into f(x) and simplify the expression. So, we get (f.g)(x) = 3(5 - x)^2 = 3(25 - 10x + x^2) = 75 - 30x + 3x^2.
Regarding your last question about the symbol "i," in terms of complex numbers, i represents the square root of -1. It is an imaginary unit that is used to solve quadratic equations with no real solutions. For example, in the equation x^2 + 1 = 0, the solutions are x = ±√(-1) = ±i.