## Factoring Trinomials – A Road Not Usually Taken

Let’s factor $10x^2+13x-3$.  Did you get $(\sqrt{10}x+\sqrt{2\imath\sqrt{30}-13}x+\imath\sqrt{3})(\sqrt{10}x-\sqrt{2\imath\sqrt{30}-13}x+\imath\sqrt{3})$?  [You can check this on Wolfram|Alpha.  Just copy and paste: (sqrt(10)x+sqrt((2*i*sqrt(30)-13)x)+i*sqrt(3))(sqrt(10)x-sqrt((2*i*sqrt(30)-13)x)+i*sqrt(3)) into the search box.] Usually we find simpler linear factors when we factor a trinomial. In our case we would get $(2x+3)(5x-1)$ or $10(x+\frac{3}{2})(x-\frac{1}{5})$.  By the way the second factorization is the best one as measured by speed and utility.  I am trying to get a paper published that makes just that point.  The factorization at the beginning of this article is ugly, weird, and wonderful.  The intent of the rest of this post is justify the “wonderful” part of that description.  To do this I will have to teach you to multiply two and three digits numbers using a nonstandard algorithm.

Consider the following graphic. (Note:  Due to my lack of knowledge about math notation and blogging, I am basically cutting and pasting from OneNote.) Multiplying Two Digit Numbers - Example

Note that the problem is worked left to right.  The hundreds are calculated first, 20 times 60=1200.  Then the tens are calculated, 20 times 5 plus 60 times 3 = 280.  I call this part cross-multiplication with addition.  Finally the ones are  3 times 5 = 15.  When I want to shake up my students as we begin a section on multiplying polynomials, I produce the right hand part of the diagram just after I ask them to multiply 23 times 65 by hand.  This is the so-called pyramid method.  You can see how it works now. Mental calculators (I am a very junior member of their club) go left to right and do the cross-multiplying with adding automatically.

Now look at this demonstration of 3 digit by 3 digit multiplication.

As before we go left to right.  The 14 stands for 200×700, 29 for 200×40+700×30, 51 for 200×2+30X40+700×5 ( all the 100’s), 26 for 30×2+40×5, and 10 for 2×5.  Notice particularly how the three multiplications are done for the 100’s.  A slight variation in the pyramid scheme is on the right.  Can you figure out what’s different and why?

Now let’s apply this method to polynomials.

The cross-multiplications with addition make the operation more efficient though it wastes space not to write the answer in one line. The diagonal has some beauty however.  Now consider this.

That is pretty cool looking.  Note we have nonstandard factorizations of $x^2+1=(x-\sqrt{2x}+1)(x+\sqrt{2x}+1)$ and $x^2-1=(x-\sqrt{2x\imath}+\imath)(x+\sqrt{2x\imath}+\imath)$.  Multiply left to right using cross-multiplication with addition and you can see where the zeroes come from.  Isn’t it really lovely how the alternating signs of the middle term in the factors produce the first and last zeroes and how the term under the square root sign is chosen to make the middle zero?

Now let’s go back to the factorization of $10x^2+13x-3$.  The steps are easier to follow if we generalize.  We wish to factor $ax^2+bx+c$ into $(\sqrt{a}x+\mu+\sqrt{c})(\sqrt{a}x-\mu+\sqrt{c})$.  We need only ensure that the middle term when we multiply is equal to $bx$, that is, $2\sqrt{ac}x-\mu^2 = bx$.  So $\mu = \sqrt{2x\sqrt{ac}-bx}$.  Our factorization is $(\sqrt{a}x+\sqrt{2x\sqrt{ac}-bx}+\sqrt{c})(\sqrt{a}x-\sqrt{2x\sqrt{ac}-bx}+\sqrt{c})$.  Applying this formula for $10x^2+13x-3$ we get $(\sqrt{10}x+\sqrt{2\imath\sqrt{30}-13}x+\imath\sqrt{3})(\sqrt{10}x-\sqrt{2\imath\sqrt{30}-13}x+\imath\sqrt{3})$.

I got interested in this structure when I was working through the very good book The Beginnings & Evolution of Algebra by Isabella Bashmakova and Galina Smirnova.  In a section on the fundamental theorem of algebra (p.95-98) the authors describe Leonard Euler’s factorization of polynomials of degree $2^n$.  For degree 4 Euler starts with $x^4+Bx^2+Cx+D=0$.  The coefficient of the $x^3$ term can always be made zero by a suitable translation.  He then shows how to factor it into the form $(x^2+\mu x+\lambda_1)(x^2-\mu x+\lambda_2)$.  I thought it was pretty cool to see the $x^3$ term disappear as I multiplied left to right.  He then used this form to find a 6th degree polynomial equation for $\mu$.  I am pretty sure Euler was an excellent mental calculator.  I guess what interested me in all this was that something as mundane as a multiplication algorithm could influence one’s view and/or discovery of a mathematical relation.

### Bonus Complication

Did you know that $10x^2+13x-3$ can be factored as $10((x+\frac{13}{20})+\sqrt{\frac{2}{20}(x+\frac{13}{20})(17\imath)}+\frac{17\imath}{20})((x+\frac{13}{20})-\sqrt{\frac{2}{20}(x+\frac{13}{20})(17\imath)}+\frac{17\imath}{20})$? [Paste 10((x+13/20)+sqrt((2/20)(x+13/20)(17i))+(17i)/20)((x+13/20)-sqrt((2/20)(x+13/20)(17i))+(17i)/20) into Wolfram|Alpha to see the factors multiplied.]  Now I know it is beginning to look like I am just adding more layers of complication but a logical progression of steps based on what we have just learned gets us there.  Again the step are easier to see with the generalization to $ax^2+bx+c=0$.

Rewrite the equation as $a(x^2+\frac{b}{a}x+\frac{c}{a})=0$.  Use the quadratic equation and the factor theorem to rewrite it as $a(x - \frac{-b+\sqrt{b^2-4ac}}{2a})(x - \frac{-b-\sqrt{b^2-4ac}}{2a})=0$.  Let $y = x-\frac{b}{2a}$ so we get $a(y-\frac{\sqrt{b^2-4ac}}{2a})(y+\frac{\sqrt{b^2-4ac}}{2a})=0$. This step is just a translation to get the roots symmetrical about zero.  This can now be rewritten as $a(y^2-(\frac{\sqrt{b^2-4ac}}{2a})^2)=0$.  Use the method we used to get the “strange” factorization of $x^2-1$ above to get $a(y+\sqrt{2y\frac{\sqrt{b^2-4ac}}{2a}}+\imath\frac{\sqrt{b^2-4ac}}{2a})(y-\sqrt{2y\frac{\sqrt{b^2-4ac}}{2a}}+\imath\frac{\sqrt{b^2-4ac}}{2a})$.  Multiply left to right with cross-multiplication and you can see those wonderful three diagonal zeroes. Now substitute $x-\frac{b}{2a}$ for $y$.  We end up with this weird, ugly and now wonderful factorization $a((x-\frac{b}{2a})+\sqrt{2(x-\frac{b}{2a})\frac{\sqrt{b^2-4ac}}{2a}}+\imath\frac{\sqrt{b^2-4ac}}{2a})((x-\frac{b}{2a})-\sqrt{2(x-\frac{b}{2a})\frac{\sqrt{b^2-4ac}}{2a}}+\imath\frac{\sqrt{b^2-4ac}}{2a})$.  Plug in the coefficients from $10x^2+13x-3$ and find $10((x+\frac{13}{20})+\sqrt{\frac{2}{20}(x+\frac{13}{20})(17\imath)}+\frac{17\imath}{20})((x+\frac{13}{20})-\sqrt{\frac{2}{20}(x+\frac{13}{20})(17\imath)}+\frac{17\imath}{20})$.

Advertisements ## About jrh794

I am a sixty-five year old math instructor at Southern Oregon University. I taught at the College of the Siskiyous in Weed California for twenty-six years. Prior to that I worked as a computer programmer, carpenter and in various other jobs. I graduated from Rice University in 1967 and have a MS in Operations Research from Stanford. In the past I have hand-built a stone house and taken long solo bicycle tours. Now I ride my mountain bike and play golf for recreation.
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