# Lesson 6

Rewriting Quadratic Expressions in Factored Form (Part 1)

### Problem 1

Find two numbers that satisfy the requirements. If you get stuck, try listing all the factors of the first number.

- Find two numbers that multiply to 17 and add to 18.
- Find two numbers that multiply to 20 and add to 9.
- Find two numbers that multiply to 11 and add to -12.
- Find two numbers that multiply to 36 and add to -20.

### Solution

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### Problem 2

Use the diagram to show that:

\((x+4)(x+2)\) is equivalent to \(x^2 +6x +8.\quad\)

\(x\) | \(2\) | |
---|---|---|

\(x\) | ||

\(4\) |

\((x-10)(x-3)\) is equivalent to \(x^2-13x+30\).

\(x\) | \(\text-10\) | |
---|---|---|

\(x\) | ||

\(\text-3\) |

### Solution

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### Problem 3

Select **all** expressions that are equivalent to \(x-5\).

\(x+(\text-5)\)

\(x-(\text-5)\)

\(\text-5 + x\)

\(\text-5 - x\)

\(5-x\)

\(\text-5 - (\text- x)\)

\(5+x\)

### Solution

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### Problem 4

Here are pairs of equivalent expressions—one in standard form and the other in factored form. Find the missing numbers.

- \(x^2 +\boxed{\phantom{300}}x + \boxed{\phantom{300}}\) and \((x-9)(x-3)\)
- \(x^2+12x+32\) and \((x+4)(x+ \boxed{\phantom{300}})\)
- \(x^2-12x+35\) and \((x-5)(x+ \boxed{\phantom{300}})\)
- \(x^2-9x+20\) and \((x-4)(x+ \boxed{\phantom{300}})\)

### Solution

### Problem 5

Find all the values for the variable that make each equation true.

- \(b(b-4.5)=0\)
- \((7x+14)(7x+14)=0\)
- \((2x+4)(x-4)=0\)
- \((\text-2+u)(3-u)=0\)

### Solution

### Problem 6

Lin charges $5.50 per hour to babysit. The amount of money earned, in dollars, is a function of the number of hours that she babysits.

Which of the following inputs is impossible for this function?

-1

2

5

8

### Solution

### Problem 7

Consider the function \(p(x)= \frac {x-3}{2x-6}\).

- Evaluate \(p(1)\), writing out every step.
- Evaluate \(p(3)\), writing out every step. You will run into some trouble. Describe it.
- What is a possible domain for \(p\)?

### Solution

### Problem 8

*Technology required. *When solving the equation \((2-x)(x+1)=11\), Priya graphs \(y=(2-x)(x+1)-11\) and then looks to find where the graph crosses the \(x\)-axis.

Tyler looks at her work and says that graphing is unnecessary and Priya can set up the equations \(2-x=11\) and \(x+1=11\), so the solutions are \(x=\text-9\) or \(x=10\).

- Do you agree with Tyler? If not, where is the mistake in his reasoning?
- How many solutions does the equation have? Find out by graphing Priya’s equation.