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25a2 - 80a + 64
(5a)2 = 25a2
25(a - 2)(a + 2)
5a(a(a-5)+3)
25a2
In step 4 of the special products section, the basic formula is given. It can also be found in most algebra books. (5a+4b)(25a2-20ab+16b2)
25a2-6400 The above is a perfectly correct answer to the question as posed. However, the question as written makes so little sense that I suspect this is another homework assignment that got slightly garbled in transmission. The original question was very likely : - 5a2 - 80 = 0. Solve for a. If so, the answer is a = +4 or -4.
3a2 - 16a + 5 = 3a2 - 15a - a + 5 = 3a(a -5) - 1(a -5) = (3a - 1)(a - 5)
This always involves a little trial and error for me. Here's what we know. There's going to be two brackets. The first terms in the brackets will be a factor pair of 35. (35,1 or 7,5) We'll have one plus and one minus. The second numbers will be a factor pair of 99. (99,1 or 33,3 or 11,9) Those will all combine to make +32, which means the plus total will be greater than the minus total. Set up your brackets. (5a +/- ?)(7a +/- ?) Then I try combinations until I find one that works. (5a + 11)(7a - 9)