Sometimes a trinomial is considered to be a special case trinomial due to the numbers involved in the coefficients. Here we focus on perfect square numbers. The table below outlines the perfect square numbers in red on the main diagonal: $$\begin{array}{|c|ccccccccccccccc|} \cdot & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ \hline 1 & \color{red}{1} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ 2 & 2 & \color{red}{4} & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & 22 & 24 & 26 & 28 & 30 \\ 3 & 3 & 6 & \color{red}{9} & 12 & 15 & 18 & 21 & 24 & 27 & 30 & 33 & 36 & 39 & 42 & 45 \\ 4 & 4 & 8 & 12 & \color{red}{16} & 20 & 24 & 28 & 32 & 36 & 40 & 44 & 48 & 52 & 56 & 60 \\ 5 & 5 & 10 & 15 & 20 & \color{red}{25} & 30 & 35 & 40 & 45 & 50 & 55 & 60 & 65 & 70 & 75 \\ 6 & 6 & 12 & 18 & 24 & 30 & \color{red}{36} & 42 & 48 & 54 & 60 & 66 & 72 & 78 & 84 & 90 \\ 7 & 7 & 14 & 21 & 28 & 35 & 42 & \color{red}{49} & 56 & 63 & 70 & 77 & 84 & 91 & 98 & 105 \\ 8 & 8 & 16 & 24 & 32 & 40 & 48 & 56 & \color{red}{64} & 72 & 80 & 88 & 96 & 104 & 112 & 120 \\ 9 & 9 & 18 & 27 & 36 & 45 & 54 & 63 & 72 & \color{red}{81} & 90 & 99 & 108 & 117 & 126 & 135 \\ 10 & 10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & \color{red}{100} & 110 & 120 & 130 & 140 & 150 \\ 11 & 11 & 22 & 33 & 44 & 55 & 66 & 77 & 88 & 99 & 110 & \color{red}{121} & 132 & 143 & 154 & 165 \\ 12 & 12 & 24 & 36 & 48 & 60 & 72 & 84 & 96 & 108 & 120 & 132 & \color{red}{144} & 156 & 168 & 180 \\ 13 & 13 & 26 & 39 & 52 & 65 & 78 & 91 & 104 & 117 & 130 & 143 & 156 & \color{red}{169} & 182 & 195 \\ 14 & 14 & 28 & 42 & 56 & 70 & 84 & 98 & 112 & 126 & 140 & 154 & 168 & 182 & \color{red}{196} & 210 \\ 15 & 15 & 30 & 45 & 60 & 75 & 90 & 105 & 120 & 135 & 150 & 165 & 180 & 195 & 210 & \color{red}{225}\\ \end{array}$$
A perfect square trinomial is any trinomial that is a result of the square of a binomial: \(ax^2+bx+c=\left(rx+s\right)^2\). For example:
| Perfect square trinomial | Non-perfect square trinomial |
|---|---|
| \(4m^2+20m+25\) | \(9p^2-8p+64\) |
| \(= \left(2m+5\right)^2\) | \(\ne \left(3p-4\right)^2\) |
There are two key points here:
| Condition #1: \(a\) and \(c\) MUST BE perfect squares (red numbers from the table above) |
| Condition #2: multiplication of: \(2\left(ax\right) \left(c\right) = bx\) MUST EQUAL the middle term |
In the example of \(4m^2+20m+25\), you can determine that the conditions hold:
| Condition #1: \(4m^2=\left(2m\right)^2\) and \(25=(5)^2\) holds since 4 and 25 are perfect squares. |
| Condition #2: a bit trickier to determine: \(2\left(2m\right) \left(5\right)=20m\) is the middle term |
When checking Condition #2, this creates a shortcut for finding the correct factors. Determining \(2m\) and \(5\) leads to the factored form \(\left(2m+5\right)^2\).
Determine if \(49z^2-14z+1\) is a perfect square trinomial. If it is, then factor.
| Check Condition #1: | \(49=(7)^2\) and \(1=(1)^2\) are perfect squares |
| Check Condition #2: | \(2 \left(7z\right) (1)=14z\) the middle term |
| Yes it is a perfect square trinomial | |
| Factored form: | \(49z^2-14z+1=\left(7z-1\right)^2\) |
Factor the trinomial \(9t^2+48tu+64u^2\).
| Check Condition #1: | \(9=(3)^2\) and \(64=(8)^2\) are perfect squares |
| Check Condition #2: | \(2 \left(3t\right)\left(8u\right)=48tu\) the middle term |
| Factored form: | \(9t^2+48tu+64u^2=\left(3t+8u\right)^2\) |
A difference of squares is easy to spot. There are only two terms, both MUST have perfect square coefficients and difference means subtraction. An example would look like: $$25y^2-49z^2$$ Notice the coefficients of \(25\) and \(49\) are perfect squares.
A couple of things to watch out for, those pesky GCFs! The following trinomial: $$108x^2-300$$ certainly does not have perfect square coefficients. However, the GCF is \(3\). Once the GCF is factored out, the binomial left over is a perfect square trinomial. $$3\cdot \left(36x^2-100\right)$$
Once the perfect square factors have been determined, those are the needed factors for the binomials and the factored form. Since there are only two terms, the middle term is missing. This happens because on of the binomials is positive and the other is negative. This causes the middle term to cancel out.
Factor the binomial: \(64r^2-81\).
| Determine the perfect square factors: | \(\implies64=\left(8\right)^2\) and \(81=\left(9\right)^2\) |
| One positive binomial, one negative: | \(\implies\left(8r-9\right)\left(8r+9\right)\) |
Every problem here can be factored using the normal procedures discussed in the last topic. The special factoring formulas presented here are for shortcut purposes.