Is there a method to find which numbers to use when simplifying quadratics?

For example $x^2 + 5x + 6$ is easy enough to find, but what if I have bigger numbers, or I have this quadratic expression: $x^2 + 7x – 6$, then it gets very hard to think of a valid simplification.

EDIT: Here's an example: $(x^2 + 5x + 6)/(x + 2)$. When simplifying the first half we get $(x + 2)(x + 3)/(x + 2)$, then we can cancel out $(x + 2)$ and the answer would be $(x + 3)$. The question is: How to always find the right numbers which to put in? (in this example they where 2 and 3)

• Do you want clever simplifications or do you accept a general algebraic property of the quadratic polynomial $$ax^2+bx+c=a(x-x_+)(x-x_-),$$ where $$x_{\pm}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\ ?$$ Jun 23 '11 at 14:09
• @AmericoTavares I don't want to solve for x, I want to simplify the quadratic expression so I could be able to cancel them out in an equation. Jun 23 '11 at 14:21
• Please excuse me. I thought you wanted to factor $ax^2+bx+c$. What do you mean by "a method to find which numbers to use when simplifying quadratics"? You can add and subtract the terms with the same degree only. Example: $3x^2+2+5x-2x^2+4=x^2+5x+6$. Jun 23 '11 at 14:36
• Correction to my first comment: Do you want clever factorizations or do you accept a general algebraic property of the quadratic polynomial ... Jun 23 '11 at 14:38
• Assuming you want to factor a quadratic polynomial with numeric coefficients, then it seems to me that what you ask is how to complete the square. Jun 23 '11 at 15:00

Suppose that you are asked to factor the polynomial $$x^2 -x-12,$$ and are unable to think of a factorization. (Yes, I know, it is easy to think of a factorization.)

You could do the following.

$1$. Solve the equation $x^2-x-12=0$ using the Quadratic Formula. In this case we find that the solutions are $$\frac{1 \pm \sqrt{(-1)^2 -(4)(1)(-12)}}{2}.$$ If you do the calculations, it turns out that the solutions are pretty simple. They are $p$ and $q$, where $p=4$ and $q=-3$.

$2$. Since the solutions are $p$ and $q$, the quadratic factors as $$(x-p)(x-q).$$ In our case, we get $$(x-4)(x-(-3)).$$ The above expression is, for good reason, considered unattractive, and probably should be rewritten as $(x-4)(x+3)$.

Let's apply the same method to $x^2-x-13$. Using the Quadratic Formula, we find that the solutions of $x^2-x-13=0$ are $p$ and $q$, where $$p=\frac{1+\sqrt{53}}{2} \qquad\text{and} \qquad q=\frac{1-\sqrt{53}}{2}.$$ So our quadratic factors as $(x-p)(x-q)$.

The general idea can be adapted to more complicated quadratics, the ones with $a\ne 1$ that are such effective instruments of torture in Algebra $1$.

Suppose that we want to factor $6x^2+x -12$. The solutions of the equation $6x^2+x-12=0$ are, by the Quadratic Formula, $p$ and $q$, where $p=16/12=4/3$ and $q=-18/12=-3/2$. Since the lead coefficient of our quadratic is $6$, we have $$6x^2+x-12=6(x-p)(x-q)=6(x-4/3)(x+3/2).$$ This looks pretty good to me, but you might want to note that we can simplify (?) this to $(3x-4)(2x+3)$.

$1$. In Algebra $1$, one is asked to acquire a facility with factoring. Many students have a good "eye", or a good memory, so experience no difficulty. (The number of examples of quadratics with smallish integer coefficients which have rational roots is quite small.)

$2$. But some find the process painful. There is a very useful formula (the Quadratic Formula) that makes the process mechanical and universal. There is an even more important idea (Completing the Square) which also makes the process mechanical. But the Quadratic Formula and Completing the Square are kept hidden. (I have some sympathy with not giving out the Quadratic Formula, and insisting that for a while all calculations be done by completing the square.)

$3$. Certain basic factoring facts, by far the most important of which are $x^2-a^2=(x-a)(x+a)$ and $x^2+bx=x(x+b)$, really are needed. But factoring $x^2-6x-91$ is only of use in quizzes.

$4$. There are real uses for factoring a quadratic as $(x-p)(x-q)$ where $p$ and $q$ may turn out to be irrational, which is most of the time. So focusing so much attention on the unusual quadratics for which $p$ and $q$ turn out to be rational is misleading.

• +1, very good. Why is the "completing the square" technique so important today? At least comparing with the 1960's, in Portugal. It seems that "the process mechanical and universal" is devaluated to an extent I am not able to understand. Jun 23 '11 at 15:33
• @Américo Tavares: There are good reasons for teaching completing the square. (i) It is how al-Khwarizmi did it, except he completed to a real square; (ii) The process can be made visual. (iii) The idea (orthogonalization) can be vastly generalized, and is useful in many places. By the way, almost everyone proves the Quadratic Formula the "wrong" (and harder) way, by dividing first by "$a$". Jun 23 '11 at 15:51

To factor a monic quadratic $x^2 + ax + b$, look for a pair of numbers whose product is $b$ and whose sum is $a$. The reason is easy to see if we work backwards: $$(x+\alpha)(x+\beta) = x^2 + (\alpha+\beta)x + \alpha\beta.$$ This works if the quadratic factors over $\mathbb{Z}$, otherwise use the quadratic formula.

• The example was $x^2 + 7x - 6$, which doesn't have a nice factorisation. Jun 23 '11 at 14:05
• Just realized that. I had $x^2 - 7x + 6$ before I noticed the mistake. Jun 23 '11 at 14:08

Eric's answer is perhaps the easiest to understand. It is equivalent to the process known as completing the square (the equivalence coming from the fact the the quadratic formula is often derived in this way). Say, we are to factor a quadratic like $x^2+ax+b$. The trick is to write the first two terms as if they were part of the square of the binomial $x+a/2$, and adjust the constant term accordingly. The square is $(x+a/2)^2=x^2+2x(a/2)+(a/2)^2=x^2+ax+a^2/4.$ So it goes like $$x^2+ax+b=(x^2+ax+a^2/4)-a^2/4+b=\left(x+\frac{a}2\right)^2-\left(\frac{a^2-4b}4\right).$$ Here we hope that the number $(a^2-4b)/4$ is non-negative, for then we can write it in the form $(a^2-4b)/4=c^2$ for some number $c$ (= a square root). So we managed to write our polynomial as a difference of two squares, and are well placed to use the familiar rule: $$x^2+ax+b=\left(x+\frac{a}2\right)^2-\left(\frac{a^2-4b}4\right)=\left(x+\frac{a}2\right)^2-c^2=(x+\frac a2+c)(x+\frac a2-c).$$ If $(a^2-4b)/4<0$, then unfortunately no factorization is possible without using complex numbers.

A reason why I often try Anononym's approach is that even when the zeros of the quadratic are both integers, it may happen that $a/2$ is not an integer. This happens when the roots have opposite parity.

Here is an additional tip. If you are factoring $$x^2 + ax + b,$$ into $(x - r)(x -s)$, the following are true. If $b > 0$ then $r$ and $s$ have the same sign. If $b < 0$, $r$ and $s$ have opposite signs. This can be helpful in narrowing the possibilities whilst factoring.

There is always the quadratic formula, which will solve any quadratic. In particular, the roots of the second equation are $$\frac{-7+\sqrt{73}}{2},\ \ \ \frac{-7-\sqrt{73}}{2}$$ so there will not be any "nice" way to factor $x^2+7x-6$.

Here is a trick that can help: Consider $ax^2+bx+c$. What if $c$ and $a$ are either prime numbers or $1$? Then the only way to factor it over the rationals is in the form $$\left(ax \pm c\right)\left(x \pm 1\right)\ \ \text{or} \ \ \left(x \pm c\right)\left(ax \pm 1\right)$$ where we have to make a choice for the $\pm$ based on the middle coefficient. (In other words, if it can be factored without square roots, it must be in the above way)

Examples: Factor $x^2-6x-7$. Well we have $(x\pm7)(x\pm1)$. We have to choose the $7$ to be negative to get $-6$ as a middle coefficient. Hence the $1$ is positive so that the product is negative ($-7$ is the last coefficient). So it factors as $(x+1)(x-7)$.

Factor $7x^2+34x-5$. The big $34$ tells us the $7$ and $5$ should appear in different factors, so we have $(7x\pm 1)(x\pm 5)$. Now we just choose the correct signs, and find $(7x-1)(x+5)$.

Hope that helps,

To factor a quadratic term like $x^2 + 7x – 6$ you make use of Vieta's theorem which says that, if $a$ and $b$ are the solutions of the equation $$x^2+p x + q = 0$$ then $$x^2+p x + q = (x-a)(x-b)$$ This is the only way to split the polynomial in two factors.

So you first solve the equation $$x^2 + 7x – 6=0$$ which gives $a=-\frac{\sqrt{73}+7}{2}$ and $b=\frac{\sqrt{73}+7}{2}$ which gives the factorization $$x^2 + 7x – 6 = (x-\frac{\sqrt{73}+7}{2})(x-\frac{\sqrt{73}+7}{2})$$ But these factors contain numbers that are not integers. So there is no such factorization you are looking for.