



Junior Member


Sep 24, 2008, 02:18 PM


factoring and producting
Can you please tell me how I would simplify these? I obviously know the answers, just not sure how to get them..
Factor #s 1  4 completely
1) 25x^2 + 10x +1 = (5x + 1)^2
2) x^3 – 3x^2 + 4x – 12 = (x – 3) (x^2 + 4)
3) x^2 + 8x + 15 = (x + 3)(x + 5)
4) x^2 + 4x + 4 = (x + 2)^2
Find the product
5) (x 1) (x + 2) = x^2 + x – 2
6) (2x + 5) (2x – 5) = 4x^2  25



Uber Member


Sep 25, 2008, 01:43 AM


I don't quite understand what you are asking, if you know the answers. Is it that you were given the answers but don't know how to get them?



Junior Member


Sep 25, 2008, 02:02 PM


Yes, I'm just not sure how to get the answers



Uber Member


Sep 26, 2008, 02:31 AM


That's a bit complicated at first. Those kinds of problems come with practice. Ok, I'll do my best. Here we go! That's how I do it.
1) 25x^2 + 10x +1 = ?
Find the factors of the coefficient of 'x^2' :
1, 5, 25. I'll choose the factor '5' (that choice come by experience, so don't worry). Therefore, I'll have:
(5x + ?)(5x + ?)
Then, the factors of the term w/o 'x'. Factors:1
So, (5x + 1)(5x + 1) = (5x + 1)^2
Suppose I chose another factor of 25, say 1. Then I would have:
(1x + ?)(25x + ?) = (x + ?)(25x + ?)
and putting the factors of 1,
(x + 1)(25x + 1)
Notice that if I expand this expression, I'll have 25x^2 + x + 25x + 1 = 25x^2 + 26x + 1, which is not the original expression.
However if you expand the first suggested solution, you'll find that:
(5x+1)(5x+1) = 25x^2 + 5x + 5x + 1 = 25x^2 + 10x + 1.



Uber Member


Sep 26, 2008, 02:59 AM


Ok, second one is more complicated.
x^3 – 3x^2 + 4x – 12 = ?
You must find a factor of '12' and try it in the expression, until it gives '0'. Remember, if you divide a number by its factor, you don't have a remainder. The '0' indicates that the remainder is '0', so a true factor.
The factors of '12' are: 1, 2, 3, 4, 6, 12 and don't forget, 1, 2, 3, 4, 6, 12.
Replace 'x' by 1 in the expression:
(1)^3 – 3(1)^2 + 4(1) – 12 = 1  3 + 4  12 = 10.
So 1 is not the good factor. That's a trial and error method. Ok, I'll go directly to the first true factor: 3
(3)^3 – 3(3)^2 + 4(3) – 12 = 27  27 + 12  12 = 0
You see? So let's continue. We now have (x  3) [It's from x=3, x3=0, so (x3)]
(x3)(?x^2 + ?x + ?).
Divide the expression by '(x3)'. Divide first x^3 by x. You have x^2. Now, put it in the solution:
(x3)(1x^2 + ?x + ?)=(x3)(x^2 + ?x + ?).
Multiply x^2 by 3, giving 3x^2. Subtract this term from the next term, that is (3x^2)  (3x^2), giving 0. Then divide by x, so '0'.
Put again in solution:
(x3)(x^2 + 0x + ?)=(x3)(x^2 + 0 + ?)=(x3)(x^2 + ?).
Multiply 0 by 3, giving 0. Subtract this from the next term, that is 4x  0, giving 4x. Then divide by x, so '4'.
But seeing '0', you can directly divide the next term '4x' by 'x', giving 4. Put again in solution:
(x3)(x^2 + 4).
To verify, multiply the last number you got by 3 in the (x3), giving 12, which is correct from the original expression.



Uber Member


Sep 26, 2008, 03:12 AM


Third one, lots easier than the earlier one.
x^2 + 8x + 15 = ?
Factors of '1' (remember, factors of the coefficient of x^2) : 1 and factors of 15: 1, 3, 5, 15.
So, (x + ?)(x + ?)
The two unknows have to make up 15, so either 1 and 15, or 3 and 5. Let's try 1 and 15.
(x + 1)(x + 15) = x^2 + x + 15x + 15 = x^2 + 16x + 15
which is wrong. So, we continue with 3 and 5.
(x + 3)(x + 5) = x^2 + 3x + 5x + 15 = x^2 + 8x + 15
Good! So the answer is (x + 3)(x + 5)!
Note that you can also have the factors of 15 as: 1, 3, 5, 15. But here, you would have got:
(x  1)(x  15) = x^2  16x + 15 and
(x  3)(x  5) = x^2  8x + 15
which are both wrong.



Uber Member


Sep 26, 2008, 03:16 AM


Try the last one, but don't forget to say if you still have difficulties.
Expansion is easier still. The easiest method I saw up to now is:
(x  1)(x + 2) = x(x + 2) 1(x + 2) = x^2 + 2x  x  2 = x^2 + x  2
once again, try the last expansion.


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