David W. answered • 07/07/15
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The problem title, “Factoring Special products,” identifies the formula and its factors as belonging to a group of formulas and factors that are “special” because they are unique and memorable (so, start to memorize them).
The difference of squares is important and easy. It looks like this: x^2 – y^2 and it is factored this way: (x+y)(x-y). To learn how this works, start with (x+y)(x-y), use F-O-I-L, then notice how the “O” and “I” terns cancel out. From now on, you don’t have to spend time using F-O-I-L to expand this --- and --- you won’t have to slow down when you factor x^2 – y^2. Later math will use this a lot, so memorizing it now will certainly save you time later.
Now, you already know that a square with side s has an area of s^2 (did you memorize that?). That means that both the larger square and the smaller square in this problem have area calculated this way. Now, think--- If the larger square has sides of length a and an area of a^2, and the “difference in the areas of the squares is a^2 – 16,” then 16 must be the value of b^2 and the value of b must be 4.
The larger square has an area of 36b^2+60b+25, which represents side^2, then each side in terms of b = (6b+5) because (6b+5)(6b+5) = 36b^2 + 60b+25 (use F-O-I-L to check this).
For the larger square, the side length is (6b+5) = 29 [since b=4) and the area is 36b^2+60b+25 which is 841. Oh, we could have just used 29*29 which is 841.
The difference of squares is important and easy. It looks like this: x^2 – y^2 and it is factored this way: (x+y)(x-y). To learn how this works, start with (x+y)(x-y), use F-O-I-L, then notice how the “O” and “I” terns cancel out. From now on, you don’t have to spend time using F-O-I-L to expand this --- and --- you won’t have to slow down when you factor x^2 – y^2. Later math will use this a lot, so memorizing it now will certainly save you time later.
Now, you already know that a square with side s has an area of s^2 (did you memorize that?). That means that both the larger square and the smaller square in this problem have area calculated this way. Now, think--- If the larger square has sides of length a and an area of a^2, and the “difference in the areas of the squares is a^2 – 16,” then 16 must be the value of b^2 and the value of b must be 4.
The larger square has an area of 36b^2+60b+25, which represents side^2, then each side in terms of b = (6b+5) because (6b+5)(6b+5) = 36b^2 + 60b+25 (use F-O-I-L to check this).
For the larger square, the side length is (6b+5) = 29 [since b=4) and the area is 36b^2+60b+25 which is 841. Oh, we could have just used 29*29 which is 841.
