# ##\$ Perfect Square Trinomials

## What is a Perfect Square Trinomial?

A perfect square trinomial is a polynomial with three terms that can be factored into two identical expressions.

You can identify a perfect square trinomial by confirming these characteristics.

• There are three terms with a $$x^2$$, $$x$$, and a constant (or an equivalent).
• $$y^2$$, $$y$$, and a constant
• $$x^6$$, $$x^3$$, and a constant
• The $$x^2$$ term has a coefficient of 1.
• The square root of the constant is equal to half of the coefficient of the $$x$$ term.

## Examples

$x^2+8x+16=(x+4)^2$

$y^2-12y+36=(y-6)^2$

$x^8+20x^4+100=(x^4+10)^2$

## How to Factor a Perfect Square Trinomial

1. Make sure the polynomial is a perfect square trinomial.
• What is the square root of the constant?
• What is half of the coefficient of x?
• Are they the same number?
2. Plug the results of Step 2 into the difference of squares formula.

## Examples

Factor: $x^2-9$

Is $$\yellow x^2 -9$$ a difference of squares?

Yes, $$\yellow x^2$$ and $$\yellow 9$$ are perfect squares that are connected with a subtraction sign.

What are the square roots of $$\yellow x^2$$ and $$\yellow 9$$?

$\sqrt{\yellow x^2}={\yellow x}$

$\sqrt{\yellow 9}={\yellow 3}$

What is the factored form of $$\yellow x^2-9$$?

To find the factored form of $$\yellow x^2-9$$, I will substitute $$\yellow x$$ and $$\yellow 3$$ into the difference of squares formula.

${\yellow x^2}-{\yellow 9}=({\yellow x}+{\yellow 3})({\yellow x}-{\yellow 3})$

The factored form of $$\yellow x^2-9$$ is…

$\yellow (x+3)(x-3)$

Factor: $36y^2-25x^2$

Is $$\green 36y^2 -25x^2$$ a difference of squares?

Yes, $$\green 36 y^2$$ and $$\green 25x^2$$ are perfect squares that are connected with a subtraction sign.

What are the square roots of $$\green 36 y^2$$ and $$\green 25x^2$$?

$\sqrt{\green 36y^2}={\green 6y}$

$\sqrt{\green 25x^2}={\green 5x}$

What is the factored form of $$\green 36y^2-25x^2$$?

To find the factored form of $$\green 36y^2-25x^2$$, I will substitute $$\green 6y$$ and $$\green 5x$$ into the difference of squares formula.

${\green 36y^2}-{\green 25x^2}=({\green 6y}+{\green 5x})({\green 6y}-{\green 5x})$

The factored form of $$\green 36y^2-25x^2$$ is…

$\green (6y+5x)(6y-5x)$

Factor: $81x^{12}-16x^4$

Is $$\purple 81x^{12}-16x^4$$ a difference of squares?

Yes, $$\purple 81x^{12}$$ and $$\purple 16x^4$$ are perfect squares that are connected with a subtraction sign.

What are the square roots of $$\purple 81x^{12}$$ and $$\purple 8x^4$$?

$\sqrt{\purple 81x^{12}}={\purple 9x^6}$

$\sqrt{\purple 16x^4}={\purple 4x^2}$

What is the factored form of $$\purple 81x^{12}-16x^4$$?

To find the factored form of $$\purple 81x^{12}-16x^4$$, I will substitute $$\purple 9x^6$$ and $$\purple 4x^2$$ into the difference of squares formula.

${\purple 81x^{12}}-{\purple 16x^4}=({\purple 9x^6}+{\purple 4x^2})({\purple 9x^6}-{\purple 4x^2})$

The difference of squares formula tells me that the factored form of $$\purple 81x^{12}-16x^4$$ is…

$\purple(9x^6+ 4x^2)(9x^6-4x^2)$

However, this expression can be factored even further because the second factor $$\purple (9x^6-4x^2)$$ is another difference of squares that can be factored to $$\purple (3x^3+2x)(3x^3-2x)$$.

So, the fully factored form of $$\purple 81x^{12}-16x^4$$ is…

$\purple (9x^6+4x^2)(3x^3+2x)(3x^3-2x)$

## Why It Works

The difference of squares formula works because factoring “un-does” polynomial multiplication.

You can see where the formula comes from if you reverse engineer the process and multiply the sum $$(a+b)$$ and difference $$(a-b)$$ of two terms.

To understand where the formula comes from…

1. Multiply $$(a+b)(a-b)$$.
2. Undo the multiplication to find the Difference of Squares Formula.

I like using the FOIL method to multiply binomials, but you can also use the box method or the multiplication algorithm. $(a+b)(a-b)$

Multiply the FIRST terms: $({\red a})({\red a})={\red a^2}$

Multiply the OUTER terms: $({\yellow a})({\yellow -b})={\yellow -ab}$

Multiply the INNER terms: $({\green b})({\green a})={\green ab}$

Multiply the LAST terms: $({\blue b})({\blue -b})={\blue -b^2}$

When the like terms ($$\yellow -ab$$ and $$\green ab$$) are combined, they cancel each other out.

The remaining terms create a difference of squares:  $(a+b)(a-b)={\red a^2}{\blue -b^2}$

If the simplified polynomial multiplication is always a difference of squares…

$(a+b)(a-b)=a^2-b^2$

Then we can “un-do” the multiplication and write the equation backwards to find the formula to factor a difference of squares…

$a^2-b^2=(a+b)(a-b)$