Perfect Square Trinomials

Lesson Features »

Lesson Priority: Normal

Algebra One $\longrightarrow$
Intro to Quadratics $\longrightarrow$
  • Multiply squared binomials formulaically instead of needing the FOIL method, thus obtaining the answer in one step
  • Instantly recognize perfect square trinomials
  • Instantly factor perfect square trinomials
Lesson Description

When you FOIL identical binomials, you obtain a trinomial that is a perfect square. The pattern that these perfect squares follow is important in a few future techniques, and it's also a nice time saver to be able to multiply identical binomials quickly, as well as factor perfect square trinomials quickly.

Practice Problems

Practice problems and worksheet coming soon!


Redefining Perfect Squares

Although the content of this lesson mostly boils down to shortcuts for things you already know, there are a few instances where you will truly benefit by recognizing this special type of trinomial. Additionally, we see these objects frequently enough that the shortcuts here are worthwhile.What we're going to look at here is what we refer to as "perfect square trinomials". Recall that for integers, we use the words "perfect square" to describe integers such as $1$, $4$, $9$, $16$, etc. because they are numbers that have integer square roots, namely because $1^2 = 1$, $2^2 = 4$, $3^2=9$, $4^2=16$, and so on. So what does that mean for variable expressions? How could a trinomial be a perfect square?Simply stated, a trinomial is a perfect square if it comes from squaring a binomial.
Define: Perfect Square TrionomialIf a trinomial quadratic is the result of squaring a binomial, then the trinomial quadratic is a perfect square.Symbolically, if a quadratic $ax^2 + bx + c$ is the result of $(dx + e)^2$ (meaning $(dx + e) \cdot (dx + e)$) where $a$, $b$, $c$, $d$, and $e$ are real number constants, then $ax^2 + bx + c$ is a perfect square trinomial.
An example of a perfect square trinomial would be $x^2 + 10x + 25$, since $(x+5)^2 = x^2 + 10x + 25$.Note that in line with something we've mentioned quickly in the past, a "quadratic" trinomial can refer to any object with a variable term, a squared variable term, and a constant term, and so things like$$ax^{2n} + bx^n + c$$can also be called quadratic, and can also be perfect square trinomials. E.g.$$(x^3 + 2)^2 = x^6 + 4x^3 + 4$$

Shortcut-ing FOIL

Ok, so we've identified and defined what perfect square trinomials are. So what's the payoff? While familiarity with these objects will just plain make your life easier sometimes, we can leverage the fact that we are squaring an object to generalize the result (a.k.a. shortcut time!).Let's use the FOIL method to square the binomial $(ax + b)$ to obtain a perfect square trinomial.$$(ax+b)(ax+b)$$$$=a^2 x^2 + abx + abx + b^2$$$$=a^2 x^2 + 2abx + b^2$$This means that, no matter what numbers $a$ and $b$ are, I can know the result of squaring the binomial without having to do the intermediate FOIL step. This may seem like no big deal, but again, even though this shortcut doesn't pay off frequently in algebra, you'll see that when it does, it's very helpful. Furthermore, we've very often working with binomials of the form $(x + b)$, so we don't even have to worry about the $a$ term most of the time.What should we take away from this? We should understand the result in words: "When you square a binomial, the result is each term squared, with the extra linear middle term, equal to twice times each constant times $x$ (or $2 \times a \times b \times x$).
Notice how the result of $(ax+b)^2$ is NOT $(a^2 x^2 + b^2)$. Trying to apply the exponent to each term (which we cannot do - that only works for factors, not sums) is possibly the most frequent mistake I've ever seen students make. Through the care of FOIL, we see why a middle term appears. Since, in this instance, we are trying to shortcut around FOIL, don't regress back to those type of mistakes! $(x+y)^2 \ne x^2 + y^2$ !!!!
Just because there is a shortcut now doesn't mean we can completely tune out! We still need to pay close attention to negative signs, just like we always do. If the $b$ term is negative then the middle linear term will be as well. E.g. $(x-3)^2 = x^2 -6x + 9$. However the constant will always be positive since the operation of squaring a term always yields a positive result.
You Should Know
Working with perfect square trinomials forward and backward between their trinomial form and their squared binomial form is a key skill to comfortably getting through one of students' least favorite quadratic equation solving techniques, completing the square ». Additionally, knowing the multiplicative shortcut for squaring a binomial not only saves you time on exams, but also severely reduces your propensity for making mistakes.
Example 1Perform the multiplication without using FOIL.$$(x-7)^2$$$\blacktriangleright$ Using our newfound shortcut, let's write the result as the square of each term in the binomial, with the extra middle linear $x$ term with a coefficient of twice the product of the binomial constants:$$(x-7)^2 = x^2 -14x + 49$$Again, the $-14$ came from multiplying the coefficients of $1$ and $-7$, and doubling the result.

Shortcutting Factoring

By studying these perfect square trinomial objects, not only do we get a shortcut for multiplying a binomial with itself, but we also get a shortcut for factoring. Let's re-examine a perfect square trinomial we've already seen:$$x^2 + 10x + 25$$We got this perfect square trinomial from squaring the binomial $(x+5)$. Now, using recently learned techniques from the lessons on Trinomial Quadratic Factoring Basics » and Advanced Trinomial Factoring », we can certainly start with an object like $x^2 + 10x + 25$ and figure out that it factors into the form $(x+5)(x+5)$. But again, recognizing the perfect square situation is a time-saving, mistake-deterring shortcut. Furthermore, as mentioned previously, we'll need to understand this factoring shortcut well in order to execute on the completing the square » equation solving process that we'll learn in the Algebra Two course.Let's define the relationship between the perfect square trinomial constants and its factored form, and thus describe the shortcut.
Define: Factored Perfect Square Trinomials:A perfect square trinomial of the form $x^2 + bx + c$ factors into the form$$\left(x+\frac{b}{2}\right)^2$$
More importantly, this relationship allows us to be able to quickly look at a trinomial and know whether or not it is a perfect square. This is one of the more convenient and useful payoffs of knowing about these objects.
Define: Identifying Perfect Square Trinomials:If you have a quadratic trinomial with the squared term coefficient $1$, e.g. $x^2 + bx + c$, and you can see that the constant $c$ term is equal to the square of one-half of $b$, then you have a perfect square trinomial.This is known as the "one-half middle" trick.
Note that this trick works well for quadratics that have a $1$ coefficient on the quadratic term. We'll mostly only study such quadratics.Examples 2-5Identify whether each of the following is a perfect square, without factoring.
Example 2$$x^2 + 8x + 16$$
Show solution
$\blacktriangleright$ Yes! This is a perfect square trinomial because $16$ is equal to the square of half of $8$.
Example 3$$x^2 + 3x + 9$$
Show solution
$\blacktriangleright$ This trinomial is not a perfect square trinomial, as $9$ is not the square of half of $3$ (though it is the square of $3$, so that means $x^2 + 6x + 9$ is a perfect square).
Example 4$$x^2 + 10x + 10$$
Show solution
$\blacktriangleright$ Half of the linear term is $5$, and $5$ squared is $25$. This means that the trinomial $x^2 + 10x + 25$ would be a perfect square, but not the trinomial $x^2 + 10x + 10$.
Example 5$$x^2 - 12x + 36$$
Show solution
$\blacktriangleright$ Yes! This is a perfect square trinomial because $36$ is equal to the square of half of $-12$. Note that having a negative sign in the linear term doesn't affect our thought process for the "one-half middle" trick, but we do need to be aware, since if we were to factor this trinomial, the correct factors would have negative signs in them. It's the difference between$$x^2 - 12x + 36 = (x-6)^2$$and$$x^2 + 12x + 36 = (x+6)^2$$

Put It To The Test

Make sure you can handle these practice problems without explicitly relying on FOIL or traditional factoring practices. Forever more, you'll be able to save time and potential mistakes using the knowledge of how perfect square trinomials work.Examples 6-8Without FOIL, multiply each of the following expressions.
Example 6$$(x+8)^2$$
Show solution
$\blacktriangleright$ Square each term in the binomial to get the "outside" terms, and the middle term is twice each binomial term.$$(x+8)^2 = x^2 + 16x + 64$$
Example 7$$(2x-5)^2$$
Show solution
$\blacktriangleright$ Same as the last example - square each term in the binomial to get the "outside" terms, and the middle term is twice each binomial term.$$(2x-5)^2 = 4x^2 - 20x + 25$$
Example 8$$(x-12)^2$$
Show solution
$\blacktriangleright$ Same as the prior examples - square each term in the binomial to get the "outside" terms, and the middle term is twice each binomial term.$$(x-12)^2 = x^2 -24x + 144$$
Examples 9-14For each of the following trinomials in Examples 9-14, state whether it is a perfect square. If it is, factor it.
Example 9$$x^2-10x+25$$
Show solution
$\blacktriangleright$ To check its status as a potential perfect square, determine if squaring half of the linear coefficient will give you the constant term.$$\left(\frac{-10}{2}\right)^2 = 25$$This works, indicating that we have a perfect square trinomial, and it factors using the $b/2$ number (in this case, $(-10/2)$ or $-5$).$$x^2 - 10x + 25 = (x-5)^2$$
Example 10$$x^2+18x+81$$
Show solution
$\blacktriangleright$ Once again, check to see if $(b/2)^2$ is the same as the constant term.$$\left(\frac{18}{2}\right)^2 = 81$$This works, indicating that we have a perfect square trinomial, and it factors using the $b/2$ number (in this case, $(18/2)$ or $9$).$$x^2+18x+81 = (x+9)^2$$
Example 11$$x^2-8x-16$$
Show solution
$\blacktriangleright$ Once again, check to see if $(b/2)^2$ is the same as the constant term.$$\left(\frac{-8}{2}\right)^2 = 16$$Don't get trapped! We need the constant to be $16$ in order for this trinomial to be a perfect square, not $-16$. This is not a perfect square trinomial.
Example 12$$x^2-20x+100$$
Show solution
$\blacktriangleright$ $(b/2)^2$ in this case is $(-10)^2 = 100$ which matches the constant term and indicates a perfect square trinomial.$$x^2-20x+100 = (x-10)^2$$
Example 13$$x^2-9x+18$$
Show solution
$\blacktriangleright$ While the linear coefficient ($9$) has common factors with the constant coefficient ($18$), we do not obtain $18$ when we take half of $9$ and square it, and therefore, this trinomial is not a perfect square.
Example 14$$x^2-5x+\frac{25}{4}$$
Show solution
$\blacktriangleright$ $(b/2)^2$ in this case is a fraction, which may not be our favorite thing to see when working with quadratics, but they are just numbers, and the rule we're looking for still applies. $(b/2)$ is $(-5/2)$, and when we square $(-5/2)$, we get $25/4$, which is exactly what we have. Therefore this is a perfect square trinomial, and the factored form will use the $(b/2)$ value:$$x^2-5x+\frac{25}{4} = \left(x - \frac{5}{2}\right)^2$$
Lesson Takeaways
  • Know what we mean when we say "perfect square trinomial"
  • Understand the relationship between the $(x-a)^2$ factored form and the coefficients you'll get in the trinomial result after you multiply
  • Be able to look at a trinomial and quickly know whether it is a perfect square
  • Factor perfect square trinomials instantly without having to rely on traditional factoring techniques

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