Question: Mel is clever, but you're pretty sure his method isn't safe to use in all situations. Explain why his method might miss some solutions. (Hint: Compare your answer to question 10 with your answer to question 5.
question 10:
10. Now, Mel says, you can find all the whole-number factor pairs for the constant number on the right side. Substitute the smaller factor in each pair for x. If the resulting equation is true, you've found a solution. Mel gives you a simple example: Solve x(x + 1) = 6 The factor pairs for 6 are 1 and 6 and 2 and 3. Try substituting 1 for x: 1(1 + 1) = 6 1(2) = 6 This is false, so x = 1 is not a solution. Try substituting 2 for x: 2(2 + 1) = 6 2(3) = 6 This is true, so x = 2 is a solution. Apply Mel's method to solve for x, the width. Does his method find the correct width?
and i got: x = 2
question 5 says:
If a product is equal to zero, we know at least one of the factors must be zero. And the constant factor cannot be zero. So set each binomial factor equal to 0 and solve for x, the width of your project.
i got x = -5 and x = 2
