Thread: Irreducible polynomials degree 2 general form

Hi all!
This should be a quick question.

if f is an irreducible polynomial (on the real line) of degree two, i need to show that it can be written in the form f(x)= (x - a)^2 + b^2
where a, and b lie on the real line and b is non-zero.

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so, I know that irreducible polynomials have no real solutions...
do I just assume f(x) can be written as x^2 + p*x+ q then complete the square and fudge it onto the expression above?
or do i need to use eucledian division?

many thanks

Yes, completing the square would give you (x+ p/2)^2+ q- p^2/4. Now consider three cases:
q- p^2/4= 0
q- p^2/4< 0 and
q- p^2/4> 0

Show that in the first and second cases, the polynomial can be factored with real coefficients but in the third, it cannot.

thanks
so then would i say a=p/2 and b^2 = q - p^2/4 ??

i also had to show the converse, like you suggested (another part of the qn). there I just expanded (x - a)^2 + b^2, equated to zero and used the quadratic formula, namely the discriminant, to show that f(x)=0 has no real roots.

i mean a = -p/2

first of all ANY monic real polynomial of degree 2 can be written in the form:

(x - a)² + c

(this is what "completing the square" does).

if this has a real root, then for some real x we have:

(x - a)² = -c

now if c < 0, then we get:

x - a = ±√(-c)

x = a ± √(-c), as our 2 (and hence all) roots of (x - a)² + c.

if c = 0, we get a "double root" a (our polynomial is (x - a)²).

however, if c > 0, then there is no real number x - a such that (x - a)² = -c < 0.

so if (x - a)² + c is irreducible (which for a polynomial of degree 2 means no roots), c must be positive, in which case c = b² for some real b (namely √c), that is:

our polynomial is of the form (x - a)² + b².

now the above only applies to MONIC polynomials, but any non-monic real polynomial is just a constant times a monic polynomial.

for example:

ax² + bx + c = a(x² + (b/a)x + c/a), and x² + (b/a)x + c/a is monic.

so your original statement isn't "quite true", for example, the polynomial:

4x² + 4 is irreducible over R (taking 4 out as a factor "doesn't count" because 4 is a UNIT (has a multiplicative inverse) in the field R), but cannot be put in the form you give.