How can i rewrite x^3-3x^2+81x-243 in the form of (x-d)(x-e)(x+f) where d is a real number and e and f are complex numbers of the form bi?

If is a polynomial of degree and are his roots, we can write

So, the exercise is basically asking you to find the three roots of the polynomial.

Cubic polynomials don't have a closed formula to solve them (well, they have, but it's very complicated), so we'd better use the rational root theorem.

This theorem states that the possible rational solutions of a polynomial with integer coefficients are of the form , where p is a divisor of the known term, and q is a divisor of the leading term.

So, in our case, we have to try all the divisors of 243 (with both signs). We find out that

So, x=3 is a solutions, and we can write as

And we can find the remaining polynomial by writing

The solutions of this polynomial are

This means that the solutions of are , ,

So, we can factor it as

answer: i guess 105 degrees?

step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: