Find a degree 3 polynomial having zeros 1,4 and 2 leading coefficient equal to 1

The degree 3 polynomial with the zeros {1, 4, 2} and a leading coefficient equal to 1 is:

p(x) = x^3 -7x^2 + 14x - 8

We know that for a polynomial of degree n, with a leading coefficient "a" and the zeros {x₁, x₂, ..., xₙ} can be written as:

p(x) = a*(x - x₁)*(x - x₂)*...*(x - xₙ)

Knowing that here we have a polynomial of degree n = 3, with a leading coefficient a = 1, and the zeros {1, 4, 2}

Replacing these in the above form, we get:

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

Now we can expand that to get:

p(x) = (x^2 - x - 4x + 4)*(x - 2) = (x^2 - 5x + 4)*(x - 2)

p(x) = x^3 - 5x^2 + 4x - 2x^2 + 10x - 8

p(x) = x^3 -7x^2 + 14x - 8

If you want to read more about polynomials, you can read:

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

3,750 deer

step-by-step explanation:

1,500 is 15 times larger than 100

lets make 250 15 times larger.

3750 deer