Is there more than one way to factor this? Show your answer using both words and mathematical notation? Provide an expression for your classmate to factor.

## Explain how to factor the following trinomials forms: x² + bx + c and ax² + bx + c.

# 3 Answers

This is actually a very good question to develop, if not evaluate, a student's understanding of factoring. Let's do a quick factoring lesson:

When factoring a trinomial (in general), the first thing to look for is a GCF (Greatest Common Factor) between each of the three terms in the trinomial. Think back to early middle school; the GCF of two numbers is a whole number that divides into both of the two numbers.

Example 1) The GCF of 20 and 15 is 5.

Example 2) The GCF of 24 and 18 is 6. Yes, 2 also divides into both 24 and 18, but we are looking for the LARGEST number that goes into both.

Once a GCF has been factored out (or if there is not a GCF), the next thing to look at are pairs of numbers that multiply to equal c/a (Remember: c/a is the product of the roots of a trinomial (You may not have learned this yet - tip for the future!). Also,
**if a=1, then you're just looking for pairs of numbers that multiply together to get c.**).

Now compare the pairs of numbers - which pair of numbers add up to b/a? (Remember: -b/a is the sum of the roots of a trinomial, but the positive or negative sign of the roots is opposite the number that is actually in the parenthesis when you factor. For
example: (x-4)(x+2)=0 --> The roots for this equation are x=4 and x=-2, but notice how the numbers in the parenthesis are actually -4 and +2. Also,
**if a = 1, you're just looking for a pair of numbers that add together to get b.**)

Last step is to put the numbers in the correct position in the GCF(_x ± #)(_x ± #) basic format for factoring trinomials.

So, to recap:

1. Look for GCF.

2. Identify pairs of numbers that multiply to get c/a.

3. Which pairs of numbers in Step #2 add to get b/a?

4. Plug the numbers & GCF into the right spot in the factoring format.

ax^{2} + bx + c = a(x – x_{1})(x – x_{2}) ,

where x_{1} and x_{2} are roots of equation

ax^{2} + bx + c = 0

and

- b ± √(b^{2} – 4ac)

x_{12} = ——————————

2a

Sorry, this algorithm presented by Patricia works, but not always. In fact, if you consider reduced trinomial x^2-7, then her algorithm will find GCF=1, pair of numbers that multiply to get -7 are 1 and -7 or -1 and 7, but what is b?? It is zero here. None of those pairs add up to zero. They add up to 6 or -6. A small trouble happened. Although this trinomial is, in fact, easily factored: x^2-7=(x-sqrt(7))*(x+sqrt(7)).

Basically, we have to use two common situations:

1) If a trinomial is in the form x^2-a^2, it is factored as (x-a)(x+a). Check this yourself by expanding. So if you have x^2-4, it can be written as x^2-2^2=(x-2)(x+2). What if we have x^2-5? It can be written as x^2-(sqrt(5))^2, then it is factored as (x-sqrt(5))*(x+sqrt(5)). Try x^2-169 now. Then x^2-85. If there is any factor in front of x^2, it can be factored out. Example: 2*x^2-200=2*(x^2-100)=2*(x-10)(x+10)--done!

2) A trinomial can be written as x^2+2*b*x+b^2 (Or x^2-2*b*x+b^2) I hope you realize the complete square here. That is, x^2+2*b*x+b^2=(x+b)^2=(x+b)*(x+b). Same is true if there is minus sign in front of 2*b*x term. x^2-2*b*x+b^2=(x-b)(x-b). Example: x^2-6x+9. It can be written as x^2-2*3*x+3^2=(x-3)(x-3). Another example: x^2+10x+25. Figure it out for yourself, I am sure you will do it easily now. If, again, there is any factor in front of x^2, see if the trinomial can be written in the form a^2*x^2+2*a*b*x+b^2 (or a^2*x^2-2*a*b*x+b^2). The rest is the same, it is a complete square and is factored easily.

3) Now in the most common case, when the trinomial is neither in reduced form nor in the form of complete square. Here we have to
**combine **two approaches.

a) First, make a*x^2 a square of some number. It is easy: a*x^2=(sqrt(a)*x)^2.

b) Second, let us create a term 2*sqrt(a)*x. But we have just b*x, you may say. OK, we will do this trick: b*x=2*sqrt(a)/(2*sqrt(a))*b*x. We multiplied and divided b*x by 2*sqrt(a)--we can do it. Let us then recombine the factors here: b*x=[2*sqrt(a)*x]*[b/(2*sqrt(a))].

3) Now, to form a complete square we need a third term, which will be the square of the second factor in square brackets above: [b/(2*sqrt(a)]. So we need b^2/(4a). But we have only c! No problem, we add and subtract that term, b^2/(4a), then we do not change anything and the expression will remain the same. And so we will do. We end up with b^2/(4a)+{c-b^2/(4a)}. Now let us combine everything together.

a*x^2+b*x+c=(sqrt(a)*x)^2+2*[sqrt(a)*x]*[b/(2*sqrt(a))]+b^2/(4a)+{c-b^2/(4a)}. Now we can form a complete square from the first three terms and obtain:

a*x^2+b*x+c=[sqrt(a)*x+b/(2*sqrt(a))]^2+{c-b^2/(4a)}. What is next? OK, let me re-write it a little bit differently: a*x^2+b*x+c=[sqrt(a)*x+b/(2*sqrt(a))]^2-{b^2/(4a)-c}. Do you see anything interesting now?

4) I hope you have already realized something: it is in the form of a reduced trinomial, same as in p. 1)!! We can easily factor it now.

[sqrt(a)*x+b/(2*sqrt(a))]^2-{b^2/(4a)-c}={[sqrt(a)*x+b/(2*sqrt(a))]-sqrt[b^2/(4a)-c]}*{[sqrt(a)*x+b/(2*sqrt(a))]+sqrt[b^2/(4a)-c]}. Looks ugly, right? OK, we will take sqrt(a) factor out from both multiples here, then we will obtain:

a*x^2+b*x+c=a*{x+b/(2a)-sqrt[b^2-4ac]/(2a)}*{x+b/(2a)+sqrt[b^2-4ac]/(2a)}

Our task is done here. Of course, this factoring is possible only if b^2-4ac>=0, which you already recognized as a discriminant from quadratic formula. If b^2-4ac=0, you end up with a complete square (check it yourself!).

Hope it helps!

## Comments

With all due respect, Kirill, the example that you gave does not support your statement that my algorithm does not work.

Point #1: You stated: "if you consider reduced trinomial x^2-7, then her algorithm will find GCF=1, pair of numbers that multiply to get 7 are 1 and 7 or -1 and -7, but what is b??"

If you factor out a GCF of 1, your reduced trinomial becomes 1 * (x^2 - 7). Since the 1 is considered trivial by mathematicians because it does not change the meaning of the equation, this would then be written as x^2 - 7, which is exactly where you started.

Secondly, 7 and 1 (and -7 and -1) are TWO pairs of numbers that multiply to 7, but as you pointed out, they are not the only ones, especially if you consider radical numbers.

Point #2: Yes, b = 0 in the example that you cited. You suggested that the factored version of x^2 - 7 = (x-sqrt(7)) * (x+sqrt(7)).

I would like to point out that -sqrt(7) + sqrt(7) = 0. And so your example still supports my algorithm.

Umm... Actually the math works from Patricia's response. It looks pretty bad to tear down another tutors response when it's actually right! You might want to avoid pointing fingers in the future.

The GCF is used to reduce or factor out the leading coefficient "a", while the roots of the polynomial are "c/a". In your example "c/a" = 7. What two numbers satisfy the constraints N1 * N2 = 7 (step #2 is Patricia's recap) and N1 + N2 = 0 (step #3 in Patricia's recap)? Positive sqrt(7) and negative sqrt(7), just like you got.

Please stick to answering the question in the future instead of impolitely, if not rudely, tearing down other peoples responses.

Sorry, guys, no offense meant, but what you said will most likely mislead student. I am not saying it is wrong, I said it does not work in the sense that is will be notoriously difficult to apply in some cases. True, √7 and -√7 multiply to -7. But they are not the only ones. right? And what about the trinomial x^2+ln(5)x-sin(10^100)? Will you see immediately, or, to that matter, in any foreseeable future two factors? I wouldn't. And I wouldn't even bother, because it does not even make any sense to try! Now, could you, please, present any objection against my points, not against the way I phrased my response?

Note: my last question in my last comment was referred to Fred's response. I am truly curious if he has any objections or corrections and would like to hear them. In fact, I did make a mistake in my original response. Too bad no one actually checked my answer, preferring to write personal reprimands instead.

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