Question

Which polynomial expression in factored form has2 as a leading coefficient with 0, -1, 6, and -4 as zeros?

Answer 1

Answer:

p(x) = 2*x^4 - 2*x^3 - 52*x^2 - 48*x

Step-by-step explanation:

For a polynomial p(x) with leading coefficient A, and zeros: {x₁, x₂, ...,xₙ}

The polynomial can be written in the factored form as:

P(x) = A*(x - x₁)*(x - x₂)*...*(x - xₙ)

In this case, we know:

Leading coefficient equal to 2, and the zeros are: 0, -1, 6, and -4

Then the polynomial can be written as:

p(x) = 2*(x - 0)*(x - (-1))*(x - 6)*(x - (-4))

p(x) = 2*x*(x + 1)*(x - 6)*(x + 4)

The wanted polynomial is this one, and we can expand it to get:

p(x) = (2*x^2 + 2*x)*(x - 6)*(x + 4)

p(x) = (2*x^3 - 12*x^2 + 2*x^2 - 12*x)*(x + 4)

p(x) = (2*x^3 - 10*x^2 - 12*x)*(x + 4)

p(x) = 2*x^4 + 8*x^3 - 10*x^3 - 40*x^2 - 12*x^2 - 48*x

p(x) = 2*x^4 - 2*x^3 - 52*x^2 - 48*x