fx is a third degree polynomial having only real coefficient

f(x) is a third degree polynomial having only real coefficients. It has 3, i, and -i as zeros, and the point (2, 25) lies on the graph. Find f(x). f(x) = x^3 - 3x^2 + x - 3 f(x) = 5x^3 - 15x^2 - 5x + 15 f(x) = -5x^3 + 15x^2 + 5x - 15 f(x) = -5x^3 + 15x^2 - 5x + 15 Show that the polynomial functions f(x) = 5x^4 - 10x - 12 has no real zero greater than 2. When f(x) divided by x - 2, the alternating signs in the bottom row indicate an upper bound. When f(x) divided by x + 2, the alternating signs in the bottom row indicate an upper bound. When f(x) is divided by x + 2, the bottom row with all signs nonnegative indicates an upper bound. When f(x) is divided by x - 2, the bottom row with all signs nonnegative indicates an upper bound. Consider the function defined by f(x) = 7x^3 + 2x^2 - 3x + 10. Use the intermediate value theorem to find two consecutive integers between which f has a zero. -1, 0 -2, -1 0, 1 1, 2 The functions f_1(x) = (x + 1)^3 and f_2(x) = (x - 5)^3 + 2 are graphed on the same axes. Explain how the graph of f_2 can be obtained by a translation of the graph of f_1. Shift the graph 6 units right and 2 units up. Shift the graph 6 units right and 2 units down. Shift the graph 5 units right and 2 units up. Shift the graph 5 units right and 2 units down.

Solution

zeros are x = 3 , i , -i

So, f(x) = a(x-3)(x^2 +1)

f(x) passes through (2 , 25)

25= a( 2 -3)(4 +1)

25 = a*-5 ; a= -5

So, f(x) = -5(x-3)(x^2+1) = -5( x^3 +x - 3x^2 -3)

= -5x^3 + 15x^2 -5x +15

Option d