Question

Which second-degree polynomial function f(x) has a lead coefficient of 4 and roots 5 and 2? f(x) = 4x² +7x+10 f(x)=4x²+28x+40 O f(x)=4x²-7x+10 Of(x)=4x²-28x+40​

Answer 1

The second-degree polynomial function f(x) that has a lead coefficient of 4 and roots 5 and 2 is f(x) = 4x² -28x + 40. The correct option is the last option f(x)=4x²-28x+40​

Quadratic equation

From the question, we are to determine which second-degree polynomial function f(x) has a lead coefficient of 4 and roots 5 and 2

To determine this, we will use the given roots to determine the equation

The roots are 5 and 2

Thus

x = 5 and x = 2

x - 5 = 0 and x - 2 = 0

Therefore,

(x -5)(x -2) = 0

Distributing

x(x -2) -5(x -2)  = 0

x² -2x -5x + 10 = 0

x² -7x + 10 = 0

Now, multiplying through by 4, we get

4x² -28x + 40 = 0

Thus, the function becomes f(x) = 4x² -28x + 40

Hence, the second-degree polynomial function f(x) that has a lead coefficient of 4 and roots 5 and 2 is f(x) = 4x² -28x + 40. The correct option is the last option f(x)=4x²-28x+40​

Learn more on Quadratic Equation here: brainly.com/question/19257086

#SPJ1