Write a polynomial equation of the least degree for each of the following sets of roots?

Question

A) 0, 2,-1
B) i, -i
please show your work, and if you can only answer one of the problems thats fine too.

Answer 1

Hi,

A) If an equation has roots of 0, 2, and -1, then it had to have factors that solved to x = 0, x = 2, and x = -1. If you move those constants to the other side, it gives you their factors:

x(x - 2)(x + 1)

If you multiply those together, that will be your polynomial equation.

x(x - 2)(x + 1) Foiling the 2 parentheses gives:
x(x^2 + 1x -2x -2)
x(x^2 - x - 2) Now distribute the x through the parentheses.
x^3 - x^2 - 2x

This is your polynomial. You can write the equation as

x^3 - x^2 - 2x = 0 or as y = x^3 - x^2 - 2x

If you put the second equation in your graphing calculator, you can see it has x intercepts at 0, 2, and -1.

B) If an equation has roots of i and -i, then it had to have factors that solved to x = i, and x = -i. If you move those "i" terms to the other side, it gives you their factors:

(x - i)(x + i)

If you multiply those together, you get:

x^2 + ix - ix - i^2

Combining like terms you get:

x^2 - i^2

But i^2 = -1, so replace it with -1 and simplify.

x^2 - (-1)

x^2 + 1

This is your polynomial. You can write the equation as

x^2 + 1 = 0 or as y = x^2 + 1


I hope this helps!! :-)

Answer 2

A) P( x) =x(x-2)(x+1)
B) P(x) =x^2+1=(x-i)*(x+i)

Answer 3

If ±4i are roots, then x² + sixteen is a element, because of fact (x+4i)(x-4i) = x² - sixteen i² = x² + sixteen. further, x² + 9 is a element. consequently, the polynomial could be divisible by (x² + 9)(x² + sixteen) If the polynomial has yet another root, that's going to might desire to be extra desirable degree, so your equation is: x^4 + 25 x² + a hundred and forty four.

Answer 4

A) p(x)= (x-0)(x-2)(x+1)= x (x^2 - x -2) = x^3 - x^2 - 2x

B) q(x)= (x-i)(x+i)= x^2 + 1