Polynomial Basics: Learn It 4

Special Cases of Polynomials

Special cases of polynomials are specific types of polynomial expressions that have distinct characteristics or follow certain patterns, making it easier to work with them, especially when you’re multiplying or factoring. Knowing these special cases is really helpful because it can simplify and speed up the process of solving polynomial equations and carrying out algebraic operations.

Perfect Square Trinomials

Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier. Let’s look at a few perfect square trinomials to familiarize ourselves with the form.

[latex]\begin{array}{ccc}\hfill \text{ }{\left(x+5\right)}^{2}& =& \text{ }{x}^{2}+10x+25\hfill \\ \hfill {\left(x - 3\right)}^{2}& =& \text{ }{x}^{2}-6x+9\hfill \\ \hfill {\left(4x - 1\right)}^{2}& =& 4{x}^{2}-8x+1\hfill \end{array}[/latex]

Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.

Perfect Square Trinomials

A perfect square trinomial is a type of polynomial that results from squaring a binomial. It is called a “perfect square” because it is the exact square of a binomial expression.

 

The general form of a perfect square trinomial is:

[latex](a+b)^2 = a^2+2ab+b^2[/latex]

Characteristics of perfect square trinomials:

  • First and Last Terms: The first and last terms are perfect squares of the terms in the original binomial.
  • Middle Term: The middle term is twice the product of the two terms in the binomial.
Expand [latex]{\left(3x - 8\right)}^{2}[/latex].Solution

[latex]\begin{align*} \text{Original expression:} & \quad (3x - 8)^2 \\ \text{Apply the formula } (a+b)^2 = a^2 + 2ab + b^2, & \quad \text{where } a = 3x \text{ and } b = -8: \\ \text{Square the first term (}a^2\text{):} & \quad (3x)^2 = 9x^2 \\ \text{Double the product of the two terms (}2ab\text{):} & \quad 2 \cdot (3x) \cdot (-8) = -48x \\ \text{Square the last term (}b^2\text{):} & \quad (-8)^2 = 64 \\ \text{Combine all terms:} & \quad 9x^2 - 48x + 64 \\ \end{align*}[/latex]

Expand [latex]{\left(4x - 1\right)}^{2}[/latex].

Difference of Squares

Another special product is called the difference of squares which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply [latex]\left(x+1\right)\left(x - 1\right)[/latex] using the FOIL method.

[latex]\begin{array}{ccc}\hfill \left(x+1\right)\left(x - 1\right)& =& {x}^{2}-x+x - 1\hfill \\ & =& {x}^{2}-1\hfill \end{array}[/latex]

The middle term drops out resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples.

[latex]\begin{array}{ccc}\hfill \left(x+5\right)\left(x - 5\right)& =& {x}^{2}-25\hfill \\ \hfill \left(x+11\right)\left(x - 11\right)& =& {x}^{2}-121\hfill \\ \hfill \left(2x+3\right)\left(2x - 3\right)& =& 4{x}^{2}-9\hfill \end{array}[/latex]

Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.

difference of squares

When you multiply a binomial by another binomial that contains the same terms but with opposite signs, the result is known as the difference of squares.

[latex](a+b)(a-b) = a^2 - b^2[/latex]

Is there a special form for the sum of squares?No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.
Multiply [latex]\left(9x+4\right)\left(9x - 4\right)[/latex].Solution

[latex]\begin{align*} \text{Original expression:} & \quad (9x + 4)(9x - 4) \\ \text{Apply the difference of squares formula, where } a = 9x, \text{ and } b = 4: & \\ \text{Use the formula } (a+b)(a-b) = a^2 - b^2: & \quad a^2 - b^2 \\ \text{Substitute } a = 9x \text{ and } b = 4: & \quad (9x)^2 - (4)^2 \\ \text{Simplify each term:} & \quad 81x^2 - 16 \end{align*}[/latex]