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# Polynomial Factorization - Series 2a - Revisited

It has been a while since my last post, but as promised I am back. Recently a reader of one of my blogs commented on how helpful they were in helping her explain a concept to her HS Junior. I was indeed very glad to hear. It certainly is an encouragement to see parents, not only sign their child up for tutoring but also willing to follow through the covered material as well. Expanding on the purpose of the blog page, would be a forum board where students/parents can post short questions and have tutor respond and perhaps even have their answers rated on helpfulness. This I believe would make a more connected community. I am sure some back-end IT development would be necessary to make it happen. Until then lets continue on polynomial factorization. I need to revisit trinomials of the quadratic form.

Polynomial Factorization - Series 2a - Revisited

Some quadratic forms of polynomials cannot be factored into roots that are integer values, and are often not very intuitive to solve using the unFOIL, chart or Trial-by-Error method. What do we do then. There is a method called simply 'Completing-the-Square' which works when all other methods, (with the exception of the quadratic formula) fail. Let us look at an example to illustrate what I mean and what to do.

Q) Find the roots (ie solve) the polynomial 2x^2 - 8x = 7

Rearranging the polynomial to standard form (ax^2 + bx + c), we have:

2x^2 - 8x - 7 = 0

Let's use first use the unFOIL method by looking for factors of -7 (the third term). Factors are -7*1 and 7*-1. None of these factors when added gives us -8. This does NOT mean we cannot factor the polynomial. It means that the solution is a real, non integer value. There are cases where polynomials have no solutions, we will talk about that later. For now let get back to completing the square.

STEP 1: Ensure that the coefficient of the squared term in x is 1. Currently it is 2. We do so by dividing the entire polynomial by two(both sides of the equation).

x^2 - 4x - 7/2 = 0

STEP 2: Add square of half the coefficient of the second term to both sides of equation(to keep things balanced). That is add the square of half of 4.

x^2 - 4x + (4/2)^2 -7/2= (4/2)^2
x^2 - 4x + 4 -7/2 = 4
x^2 - 4x + 4 = 4 + 7/2

STEP 3: The left side can be unFOIL to a perfect square. Take the square roots of both sides and solve for x.

(x-2)*(x-2) = 15/2
(x-2)^2 = 15/2
x-2 = SQRT(15/2)
x = 2+SQRT(15/2) or 2 - SQRT(15/2)

We found two roots for the equation. They were non integer real numbers 2+SQRT(15/2), 2 - SQRT(15/2).

I hope this was helpful. The typeset may be a bit difficult to follow. However note when I use the ^ symbol I am indicating an exponent follows. The expression SQRT( ) means that the expression in the parenthesis is being square rooted.

One last thing before I go. Remember the quadratic formula. Well it is based on the method of completing the square. We can derive this formula without memorization. I will blog the proof of that next.

Later!

\$55p/h

Kirk S.