Answer:
We want a polynomial of smallest degree with rational coefficients with zeros in
,
and -3. The last root gives us the factor (x+3). Hence, our polynomial is

where
is a polynomial with rational coefficients and roots
and
. The root
gives us a factor
, but in order to obtain rational coefficients we must consider the factor
.
An analogue idea works with
. For convenience write
. This gives the factor
. Hence,

Notice that
. So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

Step-by-step explanation:
We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type
will introduce in the expression, we need to multiply by its conjugate
. Hence, we will obtain
that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.
Im pretty sure the answer is D, because if i used the right formula,

you should get this in the formula


a 2 subtracting a negative 6 makes it a positive, meaning your just adding 6 plus2, to get 8
we will draw the graph according to the given constraints
NOTE: when we draw the graph from constraints inequalities becomes equalities just to draw the graph
Given constraints are:




Now we draw the graph of given constraints using graphing calculator. Please see the attachment for the graph. Shaded region is the feasible region
Answer:
$1,800
Step-by-step explanation:
If the Smiths agree to the installment plan and pay $150 every month for 12 months, it will add up to $1,800 at the end of the year.
Answer:
I think it's D
Step-by-step explanation: