College student

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+b2 and the polynomial would be q(x-A)((x-a)2+b2), where q is a real coefficient.

p(x)=q(x-A)(x2-2ax+a2+b2)=q(x3-2ax2+a2x+b2x-Ax2+2Aax-Aa2-Ab2).

p(x)=qx3-(2a+A)qx2+(a2+b2+2Aa)qx-Aq(a2+b2).

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 x2-2Ax+A2 (with duplicated zero A), and x2-2ax+a2+b2 is the quadratic factor (with complex factors x-a+ib and x-a-ib).

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