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?

### Watch This

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
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:*difference of perfect squares*

\begin{align*}x^2-y^2=(x+y)(x-y)\end{align*}

- Perfect Square Trinomial
- The
are the result of a binomial being multiplied by itself. The two variations of the perfect square trinomial are:*perfect square trinomials*

- \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*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 7.6.