andyhaus1057 Posts: 98, Reputation: 1 Junior Member #1 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
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 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?
 andyhaus1057 Posts: 98, Reputation: 1 Junior Member #3 Sep 25, 2008, 02:02 PM

Yes, I'm just not sure how to get the answers
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #4 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.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 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, x-3=0, so (x-3)]

(x-3)(?x^2 + ?x + ?).

Divide the expression by '(x-3)'. Divide first x^3 by x. You have x^2. Now, put it in the solution:

(x-3)(1x^2 + ?x + ?)=(x-3)(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:
(x-3)(x^2 + 0x + ?)=(x-3)(x^2 + 0 + ?)=(x-3)(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:
(x-3)(x^2 + 4).

To verify, multiply the last number you got by -3 in the (x-3), giving -12, which is correct from the original expression.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #6 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.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #7 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.

 Question Tools Search this Question Search this Question: Advanced Search

## Check out some similar questions!

Factoring equation [ 4 Answers ]

Factor completely: 5x 3 + 8x 2 – 5x + 8

Factoring 2x^3-1 [ 4 Answers ]

hellooo!! I need help with this question so pleeeeease help! :confused: factor 2x^3-1 ^ means to raise the power... ;) thanks again!!

Algebra 2 factoring [ 1 Answers ]

My daughter has some homework problems that I cannot help her with. The instructions say to FACTOR: Show you work. 1. 4x squared +19x-5 2. 16x squared -81y squared 3. y to the fifth power + 32x to the fifth power Can anyone out there help with explaining this?

Factoring in pre cal [ 3 Answers ]

How do you factor ; x^4-7x^3+12x^2+4x-16