Problem 1
Think of (x+4)^2 as (x+4)(x+4)
Then let y = x+4 so that we replace the first copy of "x+4" with "y"
We then get y(x+4). This distributes to xy+4y. Afterward, we plug in y = x+4 and distribute again.
So this is what the steps look like overall
z = (x+4)^2
z = (x+4)(x+4)
z = y(x+4)
z = xy + 4y
z = x(x+4) + 4(x+4)
z = x^2 + 4x + 4x + 16
z = x^2 + 8x + 16
This means the answer is choice B.
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Problem 2
We can factor out x to go from x^2 + 4x to x(x+4) showing that choice A is one of the answers.
Choice B is not an answer because (x+x)(x+4) = 2x(x+4) = 2x^2+8x which is not the same as x^2+4x.
Choice C is a similar story because (x+2)^2 = x^2+4x+4. So we can rule out choice C as well.
Choice D is almost the same as choice A, but the terms have been swapped. So choice D is one of the answers.
Choice E is another answer because (x+2)^2-4 = x^2+4x+4-4 = x^2+4x
Overall, the three answers are choice A, choice D, choice E.
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Problem 3
We can use the trick mentioned in problem 1
z = (2x-1)(3x-1)
z = y(3x-1) ..... let y = 2x-1
z = 3xy - y
z = 3x(y) - 1(y)
z = 3x(2x-1) - 1(2x-1) .... plug in y = 2x-1
z = 6x^2-3x - 2x + 1
z = 6x^2 - 5x + 1
Answer: Choice C
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Problem 4
You can use all three techniques.
In problems 1 and 3, I used the distributive property. For instance, to go from 3x(2x-1) to 6x^2-3x, I used the distributive property.
You could also use algebra tiles as a visual way to sort out the terms.
Lastly, you can use the box method, or box diagram, to split up a box and find the sub-areas through multiplication. Then add up those sub-areas to get the grand total area.