Polynomial Equation

Question

Show that (x-m) is a not factor of f(x) = 3x⁴ +x² +5 for any real number, m. 

Answer 1

Even functions can have real zeroes.  The reason the minimum is 5 is because the coefficients on the x^2 and x^4 terms are both positive.  Thus, 3x^4 >= 0 for real x and x^2 >= 0 for real x, so 3x^4 + x^2 + 5 >= 5 for real x, and there are no real roots.

 

If the x^2 coefficient had been -8 so that the polynomial was 3x^4 - 8x^2 + 5, we could factor this as

(3x^2 -5)(x^2 - 1), we would have 4 real roots, +- sqrt(5/3), +-1

Answer 2

Since this is an even function, regardless of whether x is negative, zero or positive, f*x) > 0 in all cases.

Therefore the graph is never lower than y=5, and it cannot cross the x axis.

Therefore there are no zeros (roots of f(x) = 0).

Therefore no proposed factor (x=m) can produce a zero remainder (0 remainder if required for x-m to be a factor of f(x).