Hi, spoke,
you seem to have two separate confusions,
1) how to factor the quadratic expression, and
2) how to check the result (by expanding the parentheses)
Let me begin with the second. Suppose that you have an expresion like 5(2x+3). As you know, the multiplication by 5 "distributes", giving you 10x+15. Notice that you don't operate the 2 with the 3 together; you multiply 5 by 2 on one term, and 5 by 3 on the other.
Expanding an expression like (2x+5)(x-1) is not too different: it's like two "distributive" exercises like the above (two for the price of one). It is the same as: 2x(x-1) + 5(x-1); and then you do the two parts on their own, to obtain 2x^2 - 2x + 5x - 5, that is (now grouping similar terms), 2x^2 + 3x - 5 (the "3" comes, as you see, from adding -2 and 5).
With practice, most people would expand (2x+5)(x-1) in one go: something of the form (a1+a2)(b1+b2) expands as a1.b1 + a1.b2 + a2.b1 + a2.b2. Notice that it is always and "a" with a "b" (not two "a"s or two "b"s together), and it follows the pattern "first-with-first", "first-with-second", "second-with-first", and "second-with-second"; that is, all possible combinations of one of the terms in the left (...) with one of the terms in the right (...).
With these in mind, you should be able to see now why the book result was correct, and why yours was not.
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About actually factoring the quadratic, notice one more thing in the "distributive" exercises I just mentioned. When you distribute (2x+5)(x-1), note that each (...) has:
- one term with an "x"
- and one term without an "x"
So, when you distribute all possible combinations -- one term from the left (...), one term from the right (...) --- you will get one term on "x^2" (in this example, 2x times x); TWO terms on "x" (one is 2x times -1, the other is 5 times x; these will add together to end up with 3x) and one term without any "x" (5 times -1).
So, on a quadratic expression like 2x^2 + 3x - 5, the 2x^2 comes from the product of the two terms that had an "x"; and the -5 and the end comes from the product of the two terms which did NOT have an "x". The 3x in the middle is the SUM of two of the products, in this example 2x times -1, plus 5 times x.
When factorizing a quadratic like 2x^2 + 3x - 5, you already know that the result will have the form (2x + A)(x + B), and it's all about finding the two numbers A and B. From the exercises above when expanding parentheses, you should see that
- when multiplying A times B, they must give -5, and
- when adding A + 2B, they must give 3.
Playing a little you can come up with A=5, B=-1.
It is about "reconstructing backwards" the same steps you would do "forward" when expanding parentheses.
Hope this helps!
