For factoring quadratic equations, is there a trick to find numbers that add to the b-value and multiply to the c-value?

Question

I'm trying to find a quick way to factor quadratic equations (trinomials), and I always have trouble thinking of numbers that add to the b-value and multiply to the c-value. Is there a trick to doing this?

Answer 1

Start with factoring the $c$ value and try different factors to see if you can find two factors which add up to $b$

Note that unless you are very lucky and the roots are integers or rationals, the factoring is not an easy task and using quadratic formula is advised.

You may as well have complex roots which makes factoring even more challenging.

Answer 2

Consider the quadratic equation with non-zero integer coefficients $ax^2 + bx + c = 0$.

The common "school taught" method involves figuring out a pair of signed factors for $a = a_1a_2$, and a pair of signed factors for $c = c_1c_2$, in such a way that either $a_1c_1 + a_2c_2 = b$ or $a_1c_2 + a_2c_1 = b$. The usual visual representation of the working involves a diagonal cross or a square grid or something of the sort.

The issue with this method is that it adds a layer of complexity in that you need to factorise two separate integers in different ways and figure out how to add them together to make another integer. Quite a lot of trial and error.

When I was teaching my son, I devised a slightly easier (in my view) method that simplifies the above problem.

Transform the equation to $y^2 + by + ac = 0$. This is called a monic quadratic as it has a leading coefficient of $1$. You need to find the roots of this quadratic, and a simple division will give you the roots of the original equation.

Why it works: the transformed equation (the monic) has the same discriminant and $b$ value as the original, and that's very important. The lead coefficient only scales the roots of the original to those of the transformed equation.

So you have to now factorise the single integer $ac$ into pairs of integers that add or subtract to give $b$. While $ac$ is a larger integer, remember that you only need to find divisors up to $\lfloor \sqrt{|a||c|}\rfloor$, since every composite number $N$ has at least one factor less than or equal to $\lfloor \sqrt N\rfloor$. And if you know a few divisibility rules, and have a calculator for the rest, it's really easy.

As an added simplification, when $ac$ is positive, you need both factors negative or both factors positive. This means you need to find the sum of the absolute values of the factors that give $|b|$. On the other hand, when $ac$ is negative, you need the factors to have opposite signs, so you need the difference of the absolute values of the factors that give $|b|$.

After you find the factors, affix the proper signs, and those are the roots of the transformed equation. To get the roots of the original, simply divide those roots by $a$.

I know it all sounds really complicated, but let me illustrate with a couple of examples:

$3x^2 - 26x + 35 = 0$. Ordinarily, you'd have to independently factor $3$ and $35$ and consider signed factors then see what add up to give $-26$.

Instead, transform to $y^2 - 26y + 105 = 0$.

Here, you only need to factor $105$ and you can disregard the signs (for now).

You need to work to $\lfloor \sqrt{105} \rfloor = 10$.

Let's start. Factors that fail to work are in brackets - some of these are really easy to recognise since you can't have even factors for an odd number: $1*105, (2), 3*35, (4), 5*21, (6), 7*15, (8), (9), (10)$.

Now, the final constant term is positive, so you need the sum of absolute values to equal the absolute value of the middle coefficient, $26$. In this case, you should be able to see very quickly that it's $5$ and $21$ that you need.

Since the middle coefficient is negative, you attach a negative to each of the above factors and you get the factors of the transformed equation as $(y-5)(y-21)$, with roots $5,21$.

Hence the roots of the original equation are obtained by division by the original lead coefficient, and they are: $\frac 53, 7$.

That's all you need to do to solve an equation. If you need a proper factorisation of the original expression, there's one more step:

Write down the two factors you've got: $(x- \frac 53)(x-7)$. Now multiply by the original lead coefficient to get $3(x- \frac 53)(x-7)$. You can immediately see that you can bring the $3$ into the first factor to make all coefficients integers. So your final answer is $(3x-5)(x-7)$.

Let's take another example:

$10x^2 + x - 21 = 0$. Transform into $y^2 + y - 210 = 0$. Here, the final coefficient is negative, so you need to focus on the difference in values to get the middle coefficient. You will also note that the middle coefficient is just $1$, so the factors need to be very close together (actually, consecutive). That observation makes your life even easier.

So you need to consider factors up to $\lfloor \sqrt{210} \rfloor = 14$, but you also nee them close together, so they need to be at or around the square root value. So look at $14$ and just around it. You'll find that $210 \div 14 =15$. You need one factor positive, the other negative, and the sum to be positive (the middle coefficient is positive $1$). So the negative sign has to attach to the smaller value. Hence the transformed equation has factorisation $(y+15)(y-14) = 0$. The original factorisation is $10(x +\frac{15}{10})(x-\frac{14}{10})=0$ which simplifies to $10(x +\frac{3}{2})(x-\frac{7}{5})=0$ and finally bringing in the two factors of $10$ to make all coefficients integers: $(2x+3)(5x-7)=0$. Note (again) that some of these steps are not necessary if all you need are roots to the equation, not a full factorisation. You could immediately have written the solutions as $x = -\frac 32, \frac 75$.

If you're finding it difficult to factorise even a single number (even using the trick to work up to the floor of the square root), then this algorithm is not useful. But I think it simplifies matters in most cases when $a \neq 1$.

However, it adds no value to the case $a=1$ (when the given quadratic is already a monic, or when it can easily be reduced to a monic because $a,b,c$ all share factor(s) in common).

There is one other technique that you can look at, which is well explained in this video: A different way to solve quadratics.

To sum up, that takes the monic $x^2 + bx + c$, and finds the average of roots, which can be shown (by Vieta's formulas) to be equal to $-\frac b2$. Then, because the roots of a quadratic need to be symmetrical about this average value, you can write the roots as $(-\frac b2+u)(-\frac b2-u) = \frac{b^2}4 - u^2$, which needs to equal $c$ (again by Vieta's formulas). So you solve for $u$ and get the roots as $-\frac b2+u, -\frac b2-u$. The linked video provides a practical explanation.

Note that this method is only useful for monics. But as I have shown, you can tranform any quadratic to a monic using my method (or simply dividing everything by the lead coefficient $a$) and then employ the method in the video. So you can use a combination of methods.