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 line X. I figured it out by graphing the four lines myself and when you graph line X, the slope is 3 and the y intercept is 5.
I hope this helps! Sorry it was late.
Answer:
-4x+ 1/10
Step-by-step explanation:
−1/2x−1/5−5x+3/10+3/2x
Combine like terms
-1/2x -5x + 3/2x -1/5 + 3/10
Get a common denominator for each group
-1/2x -10/2x + 3/2x -2/10 + 3/10
-8x/2 + 1/10
Simplify
-4x + 1/10
-4x + 1/10
