Factoring Quadratic ax²+bx+c with ac>0

Factoring Quadratics: $ax^2+bx+c, ac>0$

Here, we're going to learn factor by grouping of the general form of quadratics!

The first step to factor a polynomial with $ac>0$ is to find two numbers that multiply to $ac$ and add to $b$. $a$ is the leading coefficient. Assume we have found $d$ and $e$, where $de=ac$ and $d+e=b$.

We can write the polynomial as $ax^2+dx+ex+c$ because $d+e=b$. This can be grouped into $(ax^2+dx)+(ex+c)$, and both terms share a common factor. Each grouped term shares a greatest common factor. Factoring out the GCF gives us the expression of two multiplied binomials. This is the factored form of the quadratic.

Example 1:

$$3x^2 + 10x + 3$$

Since $9 ⋅ 1 = 9 = 3 ⋅ 3$, and since $9 + 1 = 10$, we can rewrite our polynomial as follows:

$$3x^2 + 9x +1x + 3$$

Now, we factor out greatest common factors from our polynomial's first two terms and from our polynomial's last two terms:

$$3x(x + 3) +1(x + 3)$$

Finally, we factor the binomial $x + 3$ out of our expression (note that we are only able to do this because our expression consists of two terms that are both being multiplied by the binomial $x + 3$):

$$(3x + 1)(x + 3)$$

If we want to double check our answer, we can find the product of the two binomials.

Here is a graphic with another example!

Test Yourself!

Let's do another problem!

Factoring trinomials algorithmically

This algorithmic method of factoring works the first time every time if you follow the steps in order. Note: if you can first factor out a Greatest Common Factor, do so. This will make the trinomial easier to factor. Then follow these steps.

Example: factor $3x^{2}+20x+12$, where $a=3$, $b=20$, and $c=12$.

1. Draw a large $X$.

2. In the top portion of the $X$, multiply $a$ and $c$ to get $ac$; here, multiply $3$ and $12$ to get $36$.

3. In the bottom portion of the $X$, write $b$; here, $20$.

4. Find two numbers that multiply to $ac$ but add up to $b$; here, $18\cdot 2=36$ and $18+2=20$. This usually takes some trial and error.

5. Write these two numbers from step (4) on the sides of the $X$.

So far, your $X$ should look like this:

1. Next, divide both side numbers (here, $18$ and $2$) by the $a$ value (here, $3$).

1. If you can, reduce your fractions; here, you can reduce $\frac{18}{3}$ to $\frac{6}{1}$.

2. Lastly, write your fractions as binomials where the denominator becomes the coefficient of the variable, and the numerator gets added on as a constant. In this case, $\frac{6}{1}$ becomes $(x+6)$ and $\frac{2}{3}$ becomes $(3x+2)$.

3. Simplify your answer into $(x+6)(3x+2)$ (or equivalently, $(3x+2)(x+6)$).

4. Check your answer by multiplying the binomials!

(Video) Common Factor Fail

by mathman1024

Here's a really helpful video by mathman1024 that teaches you a trick to factor trinomials more quickly.

$\large{\text{Summary}}$

When factoring a trinomial that doesn't contain a common factor by using trial and error, if any of the possible combinations of binomial factors that you generate contain common factors themselves, then you know for certain that those combinations will not result in your original trinomial when FOILed out.

Now, let's take a look at an example that illustrates how we can apply this trick. Suppose that we want to factor the following trinomial:

$$2x^2+7x+6$$

First off, in order to find the first terms of this trinomial's two binomial factors, we need to find two terms that multiply to the trinomial's first term: $2x^2$. Since both factors of $2x^2+7x+6$ must contain at least one term consisting of one or more variables, there exists only one pair of terms that we can break our trinomial's $2x^2$ into: $2x$ and $x$.

We then plug those terms into two empty sets of parentheses, like so:

$$(2x\ \ \ \ \ )(x\ \ \ \ \ )$$

Next, when we examine the terms that our original trinomial contains, we see that all of those terms' signs are positive, meaning that our trinomial's factors will solely contain plus signs:

$$(2x+\ \ \ )(x+\ \ \ )$$

Finally, in order to determine the last terms of our trinomial's factors, we first must look to our trinomial's last term and write down all the pairs of terms that multiply to it. In other words, we need to list all the factors of $6$:

$$\begin{array}{cc} 1&6\\ \hline 2&3 \end{array}$$

We now know that the last terms of our trinomial's factors are either $1$ and $6$ or $2$ and $3$. However, only one of these possible pairs of last terms will give us our trinomial's middle term ($7x$) when the expression's binomial factors are multiplied together.

So, let's first try plugging $1$ and $6$ into the two sets of parentheses that we've already partially filled, like so:

$$(2x+1)(x+6)$$

If we use the FOIL method to multiply this expression's inner and outer terms, we get a $1x$ and a $12x$, and when we add those two results together, we get $13x$. Since the term $13x$ has a larger coefficient than our trinomial's middle term ($7x$), we know that this potential set of factors won't result in our original trinomial when FOILed out completely.

Next, let's try swapping the last terms of the $(2x+1)(x+6)$ expression that we just ruled out:

$$(2x+6)(x+1)$$

Here's where the trick that we briefly discussed at the beginning of this explanation comes into play. We know for certain that this new expression won't result in our trinomial when FOILed out because one of the binomials it contains (i.e. $2x+6$) has a greatest common factor (i.e. $2$). A common factor can only be present in one (or both) of a trinomial's binomial factors if a common factor was present in the original trinomial to begin with. However, our trinomial does not contain a common factor, meaning that the expression $(2x+6)(x+1)$ cannot possibly be our trinomial's factored form.

So, without having to perform any further multiplication, we can definitively say that $1$ and $6$ are not the last terms of our trinomial's factors. Therefore, we can now try plugging $2$ and $3$ into our two partially filled sets of parentheses, like so:

$$(2x+2)(x+3)$$

Again, we see that one of the two binomials that comprise the expression we just created (i.e. $2x + 2$) has a greatest common factor (i.e. $2$). This means that the expression $(2x+2)(x+3)$ also cannot possibly be our trinomial's factored form, leaving us with one final combination of binomial factors to consider:

$$(2x+3)(x+2)$$

Right away, we note that neither of this new expression's binomials contain common factors. Also, when we use the FOIL method to multiply the expression's inner and outer terms, we get the terms $3x$ and $4x$; when we add those two results together, we get $7x$, which just so happens to be our trinomial's middle term.

Together, these two facts indicate that the expression $(2x+3)(x+2)$ is indeed the factored form of our trinomial:

$$\begin{align} &2x^2+7x+6\\ =&(2x+3)(x+2) \end{align}$$

After stepping us through the process of factoring the trinomial $2x^2+7x+6$ by using trial and error (in conjunction with checking for common factors, when appropriate), the video above shows us how to employ the same technique in factoring the trinomial $6x^2+5x-6$.

Factoring Quadratics when $a \neq 1$

We're going to quickly review some basics before we dive into this type of factorization.

It can be more challenging to factor a trinomial in the form $ax^2+bx+c$ than one in the form $x^2+bx+c$ because you have a new coefficient to work with.

If you multiplied $(4x+1)(2x+8)$

Using a FOIL or table method, you would get $8x^2+34x+8$.

Since factoring is the reverse of multiplication, you have to think how to multiply to get $8x^2$. You would multiply either $4x$ and $2x$ or $8x$ and $x$. Due to this, it is trickier to factor polynomials when $ac>0$ and $a$ is not $1$ because using a trial and error method, there is more than one option for the coefficient of $x$

Follow these steps to factor a trinomial $ax^2+bx+c$

1) Check to see if there is a GCF between the terms

For example, to factor the polynomial $2x^2+16x+24$ you would first notice that you could factor out or divide each term by $2$ giving you

$2\left(x^2+8x+12\right)$

Then, it is easier to factor and you would notice that $6$ and $2$ multiply to $12$ and add up to $8$. Don't forget to leave the $2$ as a factor.

The fully factored form would be: $2\left(x+6\right)\left(x+2\right)$

If there is no common monomial factor, as there often is not, that is ok! You just always want to check.

2) Use a table with $4$ sections ($2$ by $2$ dimensions). Fill in the $x^2$ term in the top left and the constant term in the bottom right.

To factor $6x^2+7x+2$ fill in the boxes as shown below

3) Multiply a and c.

In this example, $a$ is $6$ and $c$ is $2$. So $ac = 12$.

4) Find factors of $ac$ that add up to $b$.

Since $ac = 12$ and $b = 7$,

$4$ and $3$ multiply to $12$ and add up to $7$.

5) Fill in the empty boxes putting an $x$ after the numbers, as shown.

The placement doesn't matter (can be flipped in placement and the outcome will be the same)

6) Look at each row and factor out the common factor.

$6x^2+3x$ The GCF is $3x$

$4x+2$ The GCF is $2$

Therefore: $(3x+2)$ is one of the factors

7) Divide the terms by the GCF to get the other binomial factor.

$6x^2$ divided by $3x$ is $2x$

$3x$ divided by $3x$ is $1$

$\left(2x+1\right)$ is the other factor

When fully factored, $6x^2+7x+2$ = $\left(2x+1\right)\left(3x+2\right)$

(Video) Factoring Trinomials when $a>0$

by Nicole Jurewicz

This video by Nicole Jurewicz works through the process written above.