Question

p(x) = x^4 – 2x^3 + 6x^2 – 8x + 8

Luigi Ferrari (1522-1565) found a way to express such a quartic as a difference of
squares. In this case his method leads to (x^2 – x + 3)^2 – (x + 1)^2.

(a) Multiply out this last expression to check that it really is equal to p(x).
(b) Since p(x) now has the form a^2 – b^2, factorize it.
(c) Solve the two appropriate quadratic equations to find all four roots of p(x) = 0,
listing them as pairs of conjugate complex numbers.

Answer 1

\[(x^{2}-x+3)(x^{2}-x+3) = x^{4}-2x^{3}+7x^{2}-6x+9\]
\[(x+1)(x+1) = x^{2} +2x +1\]
\[(x^{4}-2x^{3}+7x^{2}-6x+9) - (x^{2}+2x+1) = x^{4}-2x^{3}+6x^{2}-8x+8\]
\[a^{2} - b^{2} = (a-b)(a+b)\]
\[=(x^{2}-2x+2)(x^{2}+4)\]
\[x^{2} -2x +2 =0 \rightarrow x= 1\pm i\]
\[x^{2} + 4=0 \rightarrow x = \pm 2i\]