Solving for the polynomial function of least degree with
integral coefficients whose zeros are -5, 3i
We have:
x = -5
Then x + 5 = 0
Therefore one of the factors of the polynomial function is
(x + 5)
Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0
Therefore one of the factors of the polynomial function is (x^2
+ 9)
The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)
= x^3 + 9x + 5x^2 = 45
Therefore, x^3 + 5x^2 + 9x – 45 = 0
f(x) = x^3 + 5x^2 + 9x – 45
The polynomial function of least degree with integral coefficients
that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45
Answer:
x¹³ - x⁹ - 9x⁸ + 9x⁴
Explanation:
Before we begin, remember the following:
xᵃ * xᵇ = xᵃ⁺ᵇ
Now, for the given:
To answer the question, we will simply multiply each term from the first bracket by each term from the second bracket and then combine like terms to get the final expression.
This can be done as follows:
(x⁸ - x⁴)(x⁵ - 9)
x⁸(x⁵) + x⁸(-9) - x⁴(x⁵) - x⁴(-9)
x¹³ - 9x⁸ - x⁹ + 9x⁴
Rearrange in order of decreasing powers, we will end up with:
x¹³ - x⁹ - 9x⁸ + 9x⁴
Hope this helps :)
Answer:
Step-by-step explanation:
cos43=60/AC
AC=60/cos43=82.04
cos27=60/BC
BC=60/cos27=67.34
sin(90-43)/BC=sin(43+27)/AB
AB=BCsin70/sin47
AB=67.34sin70/sin47
AB=86.522
P=86.522+67.34+82.04
P=235.9m
