7.6: Special Cases of Quadratic Factorization
A box is to be designed for packaging with a side length represented by the quadratic \begin{align*}9b^2 - 64\end{align*}. If this is the most economical box, what are the dimensions?
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Khan Academy Factoring the Sum and Difference of Squares
Guidance
When factoring quadratics, there are special cases that can be factored more quickly. There are two special quadratics that you should learn to recognize:
Special Case 1 (Perfect Square Trinomial): \begin{align*}x^2 \pm 2xy + y^2=(x\pm y)^2\end{align*}
- Example: \begin{align*} x^2 + 10x +25=(x+5)^2\end{align*}
- Example: \begin{align*} 4x^2 -32x + 64=(2x-8)^2\end{align*}
Special Case 2 (Difference of Perfect Squares): \begin{align*}x^2 - y^2=(x+y)(x-y)\end{align*}
- Example: \begin{align*}25x^2 - 100=(5x+10)(5x-10)\end{align*}
- Example: \begin{align*}4x^2-25=(2x-5)(2x+5)\end{align*}
Keep in mind that you can always use the box method to do the factoring if you don't notice the problem as a special case.
Example A
Factor \begin{align*}2x^2+28x+98\end{align*}.
Solution: First, notice that there is a common factor of 2. Factor out the common factor:
\begin{align*}2x^2+28x+98=2(x^2+14x+49)\end{align*}
Next, notice that the first and last terms are both perfect squares (\begin{align*}x^2=x\cdot x\end{align*} and \begin{align*}49=7\cdot 7\end{align*}, and the middle term is 2 times the product of the roots of the other terms (\begin{align*}14x=2\cdot x\cdot 7\end{align*}). This means \begin{align*}x^2+14x+49\end{align*} is a perfect square trinomial (Special Case 1). Using the pattern:
\begin{align*}x^2+14x+49=(x+7)^2\end{align*}
Therefore, the complete factorization is \begin{align*}2x^2+28x+98=2(x+7)^2\end{align*}.
Example B
Factor \begin{align*}8a^2-24a+18\end{align*}.
Solution: First, notice that there is a common factor of 2. Factor out the common factor:
\begin{align*}8a^2-24a+18=2(4a^2-12a+9)\end{align*}
Next, notice that the first and last terms are both perfect squares and the middle term is 2 times the product of the roots of the other terms (\begin{align*}12a=2\cdot 2a\cdot 3\end{align*}). This means \begin{align*}4a^2-12a+9\end{align*} is a perfect square trinomial (Special Case 1). Because the middle term is negative, there will be a negative in the binomial. Using the pattern:
\begin{align*}4a^2-12a+9=(2a-3)^2\end{align*}
Therefore, the complete factorization is \begin{align*}8a^2-24a+18=2(2a-3)^2\end{align*}.
Example C
Factor \begin{align*}x^2-16\end{align*}.
Solution: Notice that there are no common factors. The typical middle term of the quadratic is missing and each of the terms present are perfect squares and being subtracted. This means \begin{align*}x^2-16\end{align*} is a difference of perfect squares (Special Case 2). Using the pattern:
\begin{align*}x^2-16=(x-4)(x+4)\end{align*}
Note that it would also be correct to say \begin{align*}x^2-16=(x+4)(x-4)\end{align*}. It does not matter whether you put the + version of the binomial first or the – version of the binomial first.
Concept Problem Revisited
A box is to be designed for packaging with a side length represented by the quadratic \begin{align*}9b^2 - 64\end{align*}. If this is the most economical box, what are the dimensions?
First: factor the quadratic to find the value for \begin{align*}b\end{align*}.
\begin{align*}9b^2-64\end{align*}
This is a difference of perfect squares (Special Case 2). Use that pattern:
\begin{align*}9b^2-64=(3b-8)(3b+8)\end{align*}
To finish this problem we need to solve a quadratic equation. This idea is explored in further detail in another concept.
\begin{align*}& 9b^2-64=(3b+8)(3b-8)\\ & \qquad \qquad \quad \swarrow \qquad \qquad \searrow\\ & \ \quad 3b+8=0 \qquad \qquad 3b-8=0\\ & \ \qquad \ \ 3b=-8 \qquad \qquad \quad 3b=8\\ & \ \qquad \quad b=\frac{-8}{3} \qquad \qquad \quad \ b=\frac{8}{3}\end{align*}
The most economical box is a cube. Therefore the dimensions are \begin{align*}\frac{8}{3} \times \frac{8}{3} \times \frac{8}{3}\end{align*}
Vocabulary
- Difference of Perfect Squares
- The difference of perfect squares is a special case of a quadratic expression where there is no middle term and the two terms present are both perfect squares. The general equation for the difference of two squares is:
\begin{align*}x^2-y^2=(x+y)(x-y)\end{align*}
- Perfect Square Trinomial
- The perfect square trinomials are the result of a binomial being multiplied by itself. The two variations of the perfect square trinomial are:
- \begin{align*}(x+y)^2=x^2+2xy+y^2\end{align*}
- \begin{align*}(x-y)^2=x^2-2xy+y^2\end{align*}
Guided Practice
1. Factor completely \begin{align*}s^2-18s+81\end{align*}
2. Factor completely \begin{align*}50-98x^2\end{align*}
3. Factor completely \begin{align*}4x^2+48x+144\end{align*}
Answers:
1. This is Special Case 1. \begin{align*}s^2-18s+81=(s-9)^2\end{align*}
2. First factor out the common factor of 2. Then, it is Special Case 2. \begin{align*}50-98x^2=2(5-7x)(5+7x)\end{align*}
3. First factor out the common factor of 4. Then, it is Special Case 1. \begin{align*}4x^2+48x+144=4(x+6)^2\end{align*}
Practice
Factor each of the following:
- \begin{align*}s^2+18s+81\end{align*}
- \begin{align*}x^2+12x+36\end{align*}
- \begin{align*}y^2-14y+49\end{align*}
- \begin{align*}4a^2+20a+25\end{align*}
- \begin{align*}9s^2-48s+64\end{align*}
- \begin{align*}s^2-81\end{align*}
- \begin{align*}x^2-49\end{align*}
- \begin{align*}4t^2-25\end{align*}
- \begin{align*}25w^2-36\end{align*}
- \begin{align*}64-81a^2\end{align*}
- \begin{align*}y^2-22y+121\end{align*}
- \begin{align*}16t^2-49\end{align*}
- \begin{align*}9a^2+30a+25\end{align*}
- \begin{align*}100-25b^2\end{align*}
- \begin{align*}4s^2-28s+49\end{align*}
Difference of Squares
A difference of squares is a quadratic equation in the form .Perfect Square Trinomial
A perfect square trinomial is a quadratic expression of the form (which can be rewritten as ) or (which can be rewritten as ).Quadratic form
A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression.Image Attributions
Description
Learning Objectives
Here you'll learn to recognize two special kinds of quadratics and how to factor them quickly.