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
The answer is 20.4 you just multiply 1.70 by 12
Answer:
Try abc ÷ 2
Step-by-step explanation:
F Class = a
S Class = b
T Class = c
a + b + c = abc
Two buses = 2
abc ÷ 2
Answer:
Step-by-step explanation:
16.
RT=17.885
RS=9.741
Angle T=33°
17
AB=7.077
Angle B= 47.291°
Angle A= 42.709°
19
∠A = 180° - B - C = 1.36944 rad =
A=78.463°
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