x-intercepts of p(x) (the third degree polynomial) are found by solving f(x)=0. This would yield three zeroes, two of which could be complex. If all three zeroes are real, then p(x)=q(x-A)^{3}, where q is a real number. There can be at most two complex zeroes which are conjugates. If the complex zeroes are a+ib and a-ib (a, b are real) then the product of the factors would be (x-a-ib)(x-a+ib)=(x-a)^{2}+b^{2} and the polynomial would be q(x-A)((x-a)^{2}+b^{2}), where q is a real coefficient.

p(x)=q(x-A)(x^{2}-2ax+a^{2}+b^{2})=q(x^{3}-2ax^{2}+a^{2}x+b^{2}x-Ax^{2}+2Aax-Aa^{2}-Ab^{2}).

p(x)=qx^{3}-(2a+A)qx^{2}+(a^{2}+b^{2}+2Aa)qx-Aq(a^{2}+b^{2}).

Whichever the case, A would be a lone x-intercept.

Therefore the linear factor is x-A and, because A is the only x-intercept, A is real.

The real quadratic factor is x^{2}-2Ax+A^{2} (with duplicated zero A), and x^{2}-2ax+a^{2}+b^{2} is the quadratic factor (with complex factors x-a+ib and x-a-ib).