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:
1 hour 24 minutes
Step-by-step explanation:
49/35=1.4
In the addition problem attached, the letters H = 2, E = 3, A = 9. See the explanation below.
<h3>What is the explanation for the above?</h3>
In the problem, A depicts a single digit i.e; A < 9.
=> 4H<10
=>H<3
i.e; H should be 1 or 2.
In units column, 4E should have H in units place in the result. i.e; 1 or 2
As 4E is always an even number, H cannot be 1.
=> H = 2
If H = 2, E can be 3 or 8 for 4E to end in 2.
If E = 8, the result will be a three-digit number => E = 3
So if H = 2 and E = 3
A = 9
Therefore, the letters
H = 2,
E = 3, and
A = 9.
Learn more about addition problems:
brainly.com/question/24530307
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