Question 4 (2 points)
1 answer:
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Answer:
All the possible zeroes of the polynomial: f(x) = are ±1 , ±2 , ±4 , ±
, ±
, ±
by using rational zeroes theorem.
Step-by-step explanation:
Rational zeroes theorem gives the possible roots of polynomial f(x) by taking ratio of p and q where p is a factor of constant term and q is a factor of the leading coefficient.
The polynomial f(x) =
Find all factors (p) of the constant term.
Here we are looking for the factors of 4, which are:
±1 , ±2 and ±4
Now find all factors (q) of the coefficient of the leading term
we are looking for the factors of 3, which are:
±1 and ±3
List all possible combinations of ± as the possible zeros of the polynomial.
Thus, we have ±1 , ±2 , ±4 , ± , ±
, ±
as the possible zeros of the polynomial
Simplify the list to remove and repeated elements.
All the possible zeroes of the polynomial: f(x) = are ±1 , ±2 , ±4 , ±
, ±
, ±
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Givens
y = 2
x = 1
z(the hypotenuse) = √(2^2 + 1^2) = √5
Cos(u) = x value / hypotenuse = 1/√5
Sin(u) = y value / hypotenuse = 2/√5
Solve for sin2u
Sin(2u) = 2*sin(u)*cos(u)
Sin(2u) = 2() = 4/5
Solve for cos(2u)
cos(2u) = - sqrt(1 - sin^2(2u))
Cos(2u) = - sqrt(1 - (4/5)^2 )
Cos(2u) = -sqrt(1 - 16/25)
cos(2u) = -sqrt(9/25)
cos(2u) = -3/5
Solve for Tan(2u)
tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3
Notes
One: Notice that you would normally rationalize the denominator, but you don't have to in this case. The formulas are such that they perform the rationalizations themselves.
Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)
Three: If you are uncomfortable with the tan, you could do fractions.