0
$\begingroup$

I've been reading two algebra textbooks from the early 1900's. Each presents an algorithm for factoring quadratic trinomials when the coefficient of the $x^2$ term does not equal one. The technique converts the problem into one that can be solved in the same manner as solving problems where the coefficient of the $x^2$ term is equal to one (i.e., find factors of $c$ that sum to $b$).

In the first example, the coefficient of the $x^2$ term is a square and its square root is exactly contained in the coefficient of the $x$ term. Here is an example:

enter image description here

I cannot figure out why this method works and there is no additional detail in the book or anywhere else that I've looked that describes this method.

The second method applies when the $x^2$ coefficient is not a square, or is a square but that is not contained exactly in the coefficient of the $x$ term. Here is an example:

enter image description here

Conceptually, I understand how this works because the factors of $b$ should contain the same factors in $a$ and $c$, so finding factors based on a $c$ term that is the product of $a$ and $c$ makes sense. However, I don't see this connection in the first example even though in any quadratic trinomial, the following should be true: $$ (Px + m)(Qx + n) = PQx^2 + (Pn + mQ)x + mn $$

As you can see, the $b$ term contains the factors $P,n,m,Q$ which are the same factors in the $a$ term and the $c$ term.

Can anyone provide a hint as to why the first example works?

$\endgroup$
6
  • $\begingroup$ I should point out that the second example is very similar to the "grouping method" that is taught, which also multiplies "a" and "c." The second method I describe seems easier to me because I can see the logic in the algorithm, whereas the grouping method seems to make up steps that are not intuitive. $\endgroup$ Commented Jan 25, 2023 at 20:16
  • 1
    $\begingroup$ Please do not rely on pictures of text. $\endgroup$ Commented Jan 25, 2023 at 20:18
  • 2
    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ Commented Jan 25, 2023 at 20:18
  • 1
    $\begingroup$ Thank you. I will use this the next time I ask a question. $\endgroup$ Commented Jan 25, 2023 at 20:22
  • 1
    $\begingroup$ The second method is very similar to the crucial idea in deriving the quadratic formula without dealing with fractions until the end, hence it works over the integers (or any other ring where division is fraught): \begin{align} ax^2 + bx + c &= 0 \\ 4a^2x^2 + 4abx + 4ac &= 0 \\ 4a^2x^2 + 4abx + b^2 &= b^2 - 4ac \\ (2ax + b)^2 &= b^2 - 4ac \\ 2ax + b &= \pm \sqrt{b^2 - 4ac} \\ 2ax &= -b \pm \sqrt{b^2 - 4ac} \end{align} Up until this point, the process only uses addition, subtraction, multiplication, and taking a square root. Now, if you want $x$, you have to divide by $2a$. $\endgroup$ Commented Jan 25, 2023 at 22:31

1 Answer 1

Reset to default
2
$\begingroup$

In the first case you take an equation of the form $a^2 x^2+a b x+c=0$ and rewrite it as $(ax)^2+b(ax)+c=0$; the substitution $y=ax$ then yields $y^2+by+c=0$. If we instead assume the form $a^2 x^2+b x+c=0$, we first multiply by $a^2$ to get

$$a^4 x^2+a^2 b x+a^2 c=0=(a^2 x)^2+b(a^2 x)+c=0$$

So the substitution $y=a^2 x$ simplifies this to $y^2+by+c=0$. In both cases we've substituted $x=y/A$ for some $A$ to get a monic polynomial. This is typically easier to solve by inspection, since the roots $r,s$ of $$y^2+by+c=(y-r)(r-s)=0$$ will add to $b$ and multiply to $rs$. Once this is done, we back-substitute $y=Ax$ to get $$(Ax-r)(Ax-s)=0$$ which is factorized.

$\endgroup$
1
  • $\begingroup$ I still don’t understand why the first technique works. But this may be based on my understanding that the second method works because the factors of an and c are used to find b. I don’t see this relationship in the first method. $\endgroup$ Commented Jan 25, 2023 at 23:08

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.