Special Cases occur when factoring polynomials. Some may require a special pattern or process in order to be properly factored.
These cases include:
Perfect Square Trinomials are trinomials that result from a binomial multiplying (or squaring) itself (ie \(x² + 6x + 9\)).
In order for a trinomial to be classified as a perfect square, its first (\(a\)) and last (\(b\)) terms must be perfect squares (meaning their square roots must be whole numbers). In addition, the middle term must be twice as high as the product \(ab\). This can be expressed as \(\textcolor{red}{a}² + 2\textcolor{red}{a}\textcolor{blue}{b} + \textcolor{blue}{b}²\).
The form of the trinomial affects how it can be factored:
In order to factor a perfect square trinomial, we can use the Criss-Cross Method to determine \(a\) and \(b\) since it works for any trinomial. A more formal way to determine these variables would be finding the square roots of \(a²\) and \(b²\) and determine if they are whole numbers. We also need to determine if the product of \(a\) and \(b\) multiplied by \(2\) is the same as term \(2ab\) in the original expression.
Factor the trinomial \(4x² + 32x + 64\).
First, we can identify that \(a^2 = 4x^2\), \(b^2=64\), and \(2ab = 32x\).
Next, we can calculate the square roots of \(a²\) and \(b²\) to determine if they are whole numbers:
\(a = \sqrt{4x²}\)
\(\textcolor{red}{a = 2x}\)
\(b = \sqrt{64}\)
\(\textcolor{blue}{b = 8}\)
Then, we can determine \(2ab\) by finding the product of \(ab\) and multiplying it by \(2\):
\(32x = 2(\textcolor{red}{2x})(\textcolor{blue}{8})\)
\(32x = 32x\)
Since all of the values match, we can now find the factor using the first formula:
Therefore, we can determine that \(4x² + 32x + 64\) factored is \(\boldsymbol{(2x + 8)²}\).
\(4x²-12xy+9y²\)
First, we can identify that \(a^2 = 4x^2\), \(b^2 = 9y^2\), and \(-12xy\).
Next, we can determine the square roots of \(a^2\) and \(b^2\) to determine if they are whole numbers:
\(a = \sqrt{4x²}\)
\(\textcolor{red}{a = 2x}\)
\(\textcolor{blue}{b = \sqrt{9y²}}\)
\(b = 3y\)
Next, we can determine \(-2ab\) by finding the product of \(ab\) and multiplying it by \(-2\):
\(-12xy = -2(2x)(3y)\)
\(-12xy = -12xy\)
Since all of the values match, we can now find the factor using the second formula:
Therefore, we can determine that \(4x² + 32x + 64\) factored is \(\boldsymbol{(2x - 3y)²}\).
\(16 + 49d² + 70d\)
First, we can rewrite the trinomial to better identify each term:
Next, we can identify that \(a^2 = 49d^2\), \(b^2 = 16\), and \(2ab = 70d\).
Then, we can determine the square roots of \(a^2\) and \(b^2\) to determine if they are whole numbers:
\(a = \sqrt{49d²}\)
\(\textcolor{red}{a = 7d}\)
\(b = \sqrt{16}\)
\(\textcolor{blue}{b = 4}\)
After, we can determine \(2ab\) by finding the product of \(ab\) and multiplying it by \(2\):
\(70d = 2(7d)(4)\)
\(70d \neq 56d\)
We can determine that this trinomial isn't a perfect square since \(2ab ≠ 70\). As a result, we cannot factor this trinomial any further.
\(4(3x-1)²+28(3x-1)+49\)
First, we can identify that \(a^2 = 4(3x-1)²\), \(b^2 = 49\), and \(2ab = 28(3x-1)\).
Next, we can determine the square roots of \(a^2\) and \(b^2\) to determine if they are whole numbers:
\(a = \sqrt{4(3x-1)²}\)
\(\textcolor{red}{a = 2(3x-1)}\)
\(b = \sqrt{49}\)
\(\textcolor{blue}{b = 7}\)
Then, we can determine \(2ab\) by finding the product of \(ab\) and multiplying it by \(2\):
\(28(3x-1) = 2(2(3x-1))(7)\)
\(28(3x-1) = 28(3x-1)\)
Since all of the values match, we can now find the factor using the first formula. In this instance, we can actually simplify the expression further:
\((a + b)² = [2(3x-1) + 7]²\)
\((a + b)² = (6x - 2 + 7)²\)
\((\textcolor{red}{a} + \textcolor{blue}{b})² = (\textcolor{red}{6x} + \textcolor{blue}{5})²\)
Therefore, we can determine that \(4(3x-1)²+28(3x-1)+49\) factored is \(\boldsymbol{(a + b)² = (6x + 5)²}\).
A Difference of Squares is a special product that factors a binomial being subtracted into the product of 2 binomials. It can be expressed and factored as such:
In order for a binomial to be considered a difference of squares, the two terms must be perfect squares (meaning their square roots must be whole numbers). In addition, the first term, \(a\), must be postive and the second term, \(b\), must be subtracted.
Factor the binomial \(4x²-25\)
First, we can identify that \(a^2 = 4x^2\) and \(b^2 = 25\).
Next, we can find the square roots of \(a^2\) and \(b^2\) to determine if they are whole numbers:
\(a = \sqrt{4x²}\)
\(\textcolor{red}{a = 2x}\)
\(b = \sqrt{25}\)
\(\textcolor{blue}{b = 5}\)
Then, we can place these terms into two identical binomials, with one adding to each other and the other subtracting from each other:
Therefore, we can determine that \(4x²-25\) factored is \(\boldsymbol{(2x+5)(2x-5)}\).
\(100-(x-3)²\)
First, we can identify that \(a^2 = 100\) and \(b^2 = (x-3)²\).
Next, we can determine the square roots of \(a\) and \(b\) to determine if they are whole numbers:
\(a = \sqrt{100}\)
\(\textcolor{red}{a = 10}\)
\(b = \sqrt{(x-3)²}\)
\(\textcolor{blue}{b = x-3}\)
Then, we can place these terms into two identical binomials, with one adding to each other and the other subtracting from each other. In this instance, we can actually further simplify the expression:
Finally, we can simplify this equation further by collecting like terms:
Therefore, we can determine that \(100-(x-3)²\) factored is \(\boldsymbol{(7+x)(13-x)}\).
\(49r² + 81s²\)
We can determine that this expression isn't a difference of squares as the terms are being added rather than subtracted. As a result, no factoring is required.
\(50n²-72\)
First, we can factor a common factor out of the expression:
Next, we can identify that \(a^2 = 25n^2\) and \(b^2 = 36\).
Then, we can determine the square roots of \(a^2\) and \(b^2\) to determine if they are whole numbers:
\(a = \sqrt{25n²}\)
\(\textcolor{red}{a = 5n}\)
\(b = \sqrt{36}\)
\(\textcolor{blue}{b = 6}\)
Finally, we can place these terms into \(2\) identical binomials, with one adding to each other and the other subtracting from each other:
Therefore, we can determine that \(50n²-72\) factored is \(\boldsymbol{2(5n+6)(5n-6)}\).