What is a Perfect Square Trinomial?
A perfect square trinomial is a polynomial with three terms that can be factored into two identical expressions.
\[x^2+2ax+a^2=(x+a)^2\]
\[x^2-2ax+a^2=(x-a)^2\]
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
\[x^2+8x+16=(x+4)^2\]
\[y^2-12y+36=(y-6)^2\]
\[x^8+20x^4+100=(x^4+10)^2\]
When Will I Use Perfect Square Trinomials?
How to Factor a Perfect Square Trinomial
- 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?
- 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…
- Multiply \((a+b)(a-b)\).
- 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…
\[a^2-b^2=(a+b)(a-b)\]
