please help me with these math problems!!?

Question

Problem 1: Find the polynomial f(x) of degree three that has zeroes at 1,2, and 4 such as f(0)=-16. & Problem 2: The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients. & Problem 3: Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a.

can i get some serious answers please! ones that will help me!!
-THANKS!!

Answer 1

1) f(x) = A(x-1)(x-2)(x-4) for some A. Substitute in 0 to get -16 = A*-1*-2*-4, so A = 2. So 2(x-1)(x-2)(x-4).
2) Whenever you have a polynomial with real coefficients and a+bi is a root, you also have a-bi is a root. So you know the 3 roots are 4, 3+i, 3-i. Thus f(x) = (x-4)(x-(3+i))(x-(3-i)). Multiplying those last two together, you get x^2 - x(-3-i-3+i) + (3+i)(3-i) = x^2 - 6x + (9-i^2) = x^2-6x+10. So f(x) = (x-4)(x^2-6x+10).
3) I don't think you wrote the question down properly. Expanding the left hand side gives 3x^3 + (-a-27)x^2 + (9a+42)x - 14a. You need a=5 for the first part to work, but that doesn't give what you wrote.

Answer 2

1. If I understand correctly, what you're asking for is a polynomial of the form
ax^3 + bx^2 + cx + d
where f(0) = -16, and f(1), f(2), and f(4) = 0

If so, you can start by making replacing d with -16
ax^3 + bx^2 + cx -16

Then, since the whole polynomial has to equal 0 when x is either 1, 2, or 4, you know that it has to be factorable to
z(x-1)(x-2)(x-4) = 0 (z is some constant)

All you need to do then is multiply the constants in the factors together:
-1 X -2 X -4 = -8
and then figure out what this number must be mutiplied by (z) in order to get d, which was -16
z = 2

Thus, the answer in factored form is f(x) = 2(x-1)(x-2)(x-4)
Multiply it out to get it in polynomial form.

Answer 3

Answer to problem #1:

We know that, because it has degree 3, the function is of the form:
f(x)=Ax^3+Bx^2+Cx+D

if f(0) = -16 then we know that D will have to be -16 because it is the only constant and all other terms involve a power of x and will be 0

we also know that the zeros are a 1, 2, and 4

therefore:

(x-1)*(x-2)*(x-4)=0

we multiply this out and get:

x^3-7x^2+14x-8=0

ok now multiply this whole thing by 2 to make our D term a -16 and we get:

f(x)=2x^3-14x^2+28x-16

this satisifies all of the conditions given for the function

Answer to Problem #3:
I just multiplied out, factored, solved for a, and simplified

a=(87x-151)/((x-7)*(x-2))=5

?????

Answer 4

Problem 1 / if the roots are 1 ,2,4 the plynomial ha s the form
k(x-1) (x-2) (x-4) =f (y ) where k is a constant. For x=0 you have
-8k = 16 so k = -2 and
f(y) = -2x^3 +14 x^2-28x +16

Problem 2 I do not know

Answer 5

wat math class do u have i have algabra 1 honors and we havnt doen that so prob 3 seems like something we will do though

Answer 6

im only in 6th grade!!!!!!

(learning 7th grade math and 8th grade english! :D)

Answer 7

No

Answer 8

ummm...... good luck with that

Answer 9

WHAT?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!??!?!?!?!?!??!