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

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1