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Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

• Write quadratic equations in standard form.
• Factor quadratic expressions for different coefficient values.
• Factor when a=1\begin{align*}a = -1\end{align*}.

## Vocabulary

Terms introduced in this lesson:

## Teaching Strategies and Tips

In this lesson, students learn to factor quadratic polynomials according to the signs of a,b,\begin{align*}a, b,\end{align*} and c\begin{align*}c\end{align*}:

• a=1,b>0,c>0\begin{align*}a = 1, b > 0, c > 0\end{align*}. See Examples 1-4.

Factor.

a. x2+15x+26\begin{align*}x^2 + 15x + 26\end{align*}. Answer: (x+13)(x+2)\begin{align*}(x + 13)(x + 2)\end{align*}

b. x2+13x+40\begin{align*}x^2 + 13x + 40\end{align*}. Answer: (x+8)(x+5)\begin{align*}(x + 8)(x + 5)\end{align*}

c. x2+20x+75\begin{align*}x^2 + 20x + 75\end{align*}. Answer: (x+15)(x+5)\begin{align*}(x + 15)(x + 5)\end{align*}

• a=1,b<0,c>0\begin{align*}a = 1, b < 0, c > 0\end{align*}. See Examples 5 and 6.

Factor.

a. x217x+42\begin{align*}x^2 - 17x + 42\end{align*}. Answer: (x14)(x3)\begin{align*}(x - 14)(x - 3)\end{align*}

b. x221x+90\begin{align*}x^2 - 21x + 90\end{align*}. Answer: (x15)(x6)\begin{align*}(x - 15)(x - 6)\end{align*}

c. x214x+48\begin{align*}x^2 - 14x + 48\end{align*}. Answer: (x6)(x8)\begin{align*}(x - 6)(x - 8)\end{align*}

• a=1,c<0\begin{align*}a = 1, c < 0\end{align*}. See Examples 7-9.

Factor.

a. x215x54\begin{align*}x^2 - 15x - 54\end{align*}. Answer: (x18)(x+3)\begin{align*}(x - 18)(x + 3)\end{align*}

b. x2+7x60\begin{align*}x^2 + 7x - 60\end{align*}. Answer: (x+12)(x5)\begin{align*}(x + 12)(x - 5)\end{align*}

c. x216x192\begin{align*}x^2 - 16x - 192\end{align*}. Answer: (x24)(x+8)\begin{align*}(x - 24)(x + 8)\end{align*}

• a=1\begin{align*}a = -1\end{align*}. See Example 10.

Factor.

a. x24x+60\begin{align*}-x^2 - 4x + 60\end{align*}. Answer: (x6)(x+10)\begin{align*}-(x - 6)(x + 10)\end{align*}

b. x2+14x40\begin{align*}-x^2 + 14x - 40\end{align*}. Answer: (x10)(x4)\begin{align*}-(x - 10)(x - 4)\end{align*}

c. x225x156\begin{align*}-x^2 - 25x - 156\end{align*}. Answer: (x+12)(x+13)\begin{align*}-(x + 12)(x + 13)\end{align*}

• Allow students to infer that if c>0(a=1)\begin{align*}c > 0 (a = 1)\end{align*}, then the factorization will be either of the form (+)\begin{align*}(\underline{\;\;\;\;} + \underline{\;\;\;\;})\end{align*}(+)\begin{align*}(\underline{\;\;\;\;} + \underline{\;\;\;\;})\end{align*} or ()\begin{align*}(\underline{\;\;\;\;} - \underline{\;\;\;\;})\end{align*}()\begin{align*}(\underline{\;\;\;\;} - \underline{\;\;\;\;})\end{align*} (same signs). If c<0\begin{align*}c < 0\end{align*} (a=1)\begin{align*}(a = 1)\end{align*}, then use the form ()\begin{align*}(\underline{\;\;\;\;} - \underline{\;\;\;\;})\end{align*}()\begin{align*}(\underline{\;\;\;\;} - \underline{\;\;\;\;})\end{align*} (different signs).
• See summary at the end of the lesson for a list of procedures and examples for each case.

Emphasize that factoring is the reverse of multiplication.

• Use an example such as (x+3)(x+7)=x2+10x+21\begin{align*}(x + 3)(x + 7) = x^2 + 10x + 21\end{align*} in which the binomials are expanded one step at a time to motivate factoring.
• Demonstrate that factoring is equivalent to putting squares and rectangles back together into larger rectangles.

Example:

Multiply.

(x+3)(x+7)\begin{align*}(x + 3)(x + 7)\end{align*}.

Solution. The diagram shows that (x+3)(x+7)=x2+10x+21\begin{align*}(x + 3)(x + 7) = x^2 + 10x + 21\end{align*}. Observe that it also shows how to factor x2+10x+21\begin{align*}x^2 + 10x + 21\end{align*}.

Suggest that students stop listing the possible products for c\begin{align*}c\end{align*} after the correct choice is evident.

## Error Troubleshooting

General Tip: For quadratic trinomials with a=1\begin{align*}a = -1\end{align*}, remind students to factor 1\begin{align*}-1\end{align*} from every term. Remind students to include it in their final answer.

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