If a third degree polynomial has a lone x-intercept at x =A , discuss what this implies about the linear and quadratic factors of that polynomial.

Question

College student

Answer 1

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).