Completing the square

Question

Why can't you factor a quadratic function completely all the time instead of completing the square.

Answer 1

In general, how much a polynomial can be factored, or if it factors at all, depends on the number system that you allow the coefficients of the factors to belong to.

Mathematicians call any such number system a Field, equipped with operations + and ×. You can read the precise definition on Wikipedia, but all the properties required for a Field are taught in late grade school to early middle school, though the term "Field" is not taught.

For example, rational numbers Q, real numbers R, and complex numbers C form fields.

Examples:

x² - 1 factors in Q as (x + 1)(x - 1) and hence also factors in R and C.

x² - 2 doesn't factor in Q. It does factor in R as (x + √2)(x - √2) and hence also factors in C.

Finally, x² + 1 doesn't factor in Q or R. It factors in C as (x + i)(x - i).

The fundamental theorem of algebra states that every degree n polynomial with coefficients in C, factors in C as f(x) = a(x - r₁)(x - r₂)...(x - r_n).

Answer 2

Hi Shauntia,

Factoring a quadratic equation only works when the solution has roots which are rational numbers - meaning that each root is a quotient of 2 integers. 

Otherwise, the solution can only be found by using the quadratic formula or by completing the square.  The quadratic formula is actually derived by completing the square on the standard form of the quadratic equation (ax^2 + bx + c = 0).  Below is the derivation:

1.  ax^2 + bx + c = 0 (standard form of quadratic equation)

2.  x^2 + bx/a + c/a = 0 (divide both sides by a to get rid of x^2 coefficient on left side)

3.  x^2 + bx/a = -c/a (move constant term from left side to right side by subtracting c/a from both sides)

4.  x^2 + bx/a + b^2/4a^2 = b^2/4a^2 - c/a (take half of coefficient of x term on left side and square it and add result to both sides)

5.  (x + b/2a)^2 = b^2/4a^2 - c/a (rewrite trinomial on left side as perfect binomial square)

6.  x + b/2a = +/- sqrt(b^2/4a^2 - c/a) (take square root of both sides)

7.  x + b/2a = +/- sqrt(b^2 - 4ac / 4a^2) (combine fractions under square root symbol on right side under common denominator of 4a^2)

8.  x + b/2a = +/- sqrt(b^2 - 4ac) / 2a (extract perfect square from denominator on right side)

9.  x = -b/2a +/- sqrt(b^2 - 4ac) /2a (isolate x term on left side by subtracting b/2a from both sides)

10.  Quadratic Formula:  x = (-b +/- sqrt(b^2 - 4ac)) / 2a (combine fractions on right side under common denominator of 2a) 

So we can see from the above derivation that using the quadratic formula is the equivalent of completing the square.  We must remember that using the quadratic formula or completing the square is necessary when the solution involves irrational root(s) - meaning at least of the roots cannot be expressed as a quotient of 2 integers. 

We must also remember that to complete the square - the coefficient of the x^2 term must be 1.  That is why the 1st step in the derivation of the quadratic formula via completing te square on the standard fom of the quadratic equation was to divide each term by the coefficient of the x^2 term to ensure a coefficient of 1 for that term.

I hope the above explanation helps you both to know when factoring does NOT work and also how the quadratic formula works as a result of completing the square on the standard form of the quadratic equation.

Thanks for submitting your question.

Regards, Jordan. 

Answer 3

If a quadratic function is prime, you can't factor it out in a regular way. However, no matter how complicated a quadratic function is, you can always factor it out by completing the square, the same way as when you derive the quadratic formula.

Example:

y = x^2+4x+5 is prime since b^2-4ac = -4 is not a perfect square.

By completing the square,

y = (x+2)^2 + 1 = (x+2+i)(x+2-i)