P(x)=x^3-6x^2+2x-12 has two complex roots and one real root as seen on the graph. Show that i radical 2 is a complex root of P(x). Hint: remember that i^2=-1.

Answer 1

P(x)=x^3-6x^2+2x-12 has two complex roots and one real root as seen on the graph. Show that i radical 2 is a complex root of P(x). Hint: remember that i^2=-1.

Roots means: x values that makes P(x) = 0.

P(i√2) = (i√2)^3 -6 (i√2)^2 + 2 * i√2 -12

[i^2 = -1 , i ^ 3 = -i ]

=-i * 2√2 - 6 * (-1 * 2) + 2i√2 - 12

= -2i√2 + 2i√2 -12 + 12

= 0

Since P(i√2) = 0, i√2 is root of P(X).

Answer 2

A graph was not included, but I graphed the function and determined that the only real root is x = 6. This means that ( x - 6 ) is a factor of P(x).

You can use either polynomial long division or synthetic division to factor P(x). If you need help with this, please post a comment on this answer, and I will be glad to show you how this is done.

The result is x² + 2. Luckily, you won't need the quadratic equation to factor this.

Set x² + 2 = 0 and then subtract 2 from both sides of the equation.

Thus x² = -2. Taking the square root of both sides of this equation results in x = √( - 2 ) = √[ 2 * ( - 1 ) ]

= ( √2 ) * √( -1) = ( √2 ) * i = i √2

AND, by the way, the third and final root is -i √2.

If the coefficients of a polynomial are all real numbers, then any complex roots come in conjugate pairs, which have the form: a + bi and a - bi. In this case, a = 0. ( a and b are real numbers.)