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**SlowlyFading****Member**- Registered: 2012-06-12
- Posts: 149

These are the ones I got wrong and I need the answers and work for them.

I'm just here to get some help with an online math course I'm taking.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi SlowlyFading

Are these the problems?

And I'm not sure what 5) is supposed to be. Probably a difference of squares. Is there supposed

to be a perfect square where the ??? is?

For the subtraction symbol use the hyphen (the lower case of the key just to the right of the

") zero" key). Then the LaTex will come out looking much better. Also use the "^" for exponents.

Click on the LaTex on my post and you can see how to input the code. The "\;" just puts a space

before the first term in each problem.

The first two seem to be factoring a common factor out of each of the terms: x in 1) and 17 in 2).

The rest seem to be "difference of squares", "perfect square trinomials" or "trial and error", BUT

one of them looks like it WILL NOT factor.

I hope that helps. If not then we can give more direction.

Keep hanging in there!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**SlowlyFading****Member**- Registered: 2012-06-12
- Posts: 149

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

**Those are the ones I need the answers and work (all the steps used) for..**

I'm just here to get some help with an online math course I'm taking.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

Q2

Looks like (x+1) is also a factor.

It is, but you aren't required to show it.

Try

http://www.mathsisfun.com/algebra/factoring.html

I am still thinking about other factors for Q1

BIT LATER EDIT:

No other factors found for Q1

Bob

ps. I'd offer to take you through them one at a time, but that didn't seem to work last week.

pps. Look what I've found:

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi Slowly Fading!

Have you access to an explanation of the "FOIL" method? It can be used to do the factoring of #3 through #9. (10 doesn't factor). This method will work on perfect square trinomials such as

9x^2 - 24x + 16 = (3x-4)^2; and on trial and error types such as x^2 - 2x - 24 = (x-6)(x+4); and on

differences of squares such as 9x^2 - 4 = 9x^2 + 0x - 4 = (3x-2)(3x+2).

Let us know how it's going.

Have a great day!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi SlowlyFading!

Here's another way of looking at the factoring of ax^2 + bx + c ( a not zero) trinomials. It is fully

equivalent to the FOIL method, but has a two dimensional set up rather than one dimensional.

The FOIL technique is equivalent to RCM (reversing cross multiplication for a 2x2 multiplication).

Here is the pattern:

. . . . . . where these three patterns from RIGHT TO LEFT produce

| X | the units, 10', and 100's columns of a product in base 10.

. . . . . . In an unknown base x (polynomials) this is units, x's, x^2's.

Example:

2 -3 Obtain the 3*(-4)=-12 from the RIGHT HAND pattern.

Obtain 2*4+3*(-3) = -1 from the CROSS pattern.

3 4 Obtain 2*3=6 from the LEFT HAND pattern.

____

6 -1 -12

As polynomials this is equivalent to (2x-3)(3x+4) = 6x^2 - x - 12.

Putting this "in reverse" we would be given the 6 -1 -12 and asked to find the factors of this

which are 2 -3 and 3 4.

Ignoring the signs temporarily there are 6 possible ways to pair the factorizations ( 1*6 or 2*3 )

with the factorizations ( 1*12 or 2*6 or 3*4 ). Then we can switch the numbers in one of the

columns (but not in both at the same time) which doubles the number of possibilities to 12.

It's a matter of trial and error to find out which combination works (if any). The more practice

one has doing this, the faster they can find the right one.

HELPFUL HINT for eliminating some possibilities: If the product ( in this case the 6 -1 -12 ) has no

common factor > 1 (as is the case here) then NEITHER of the two factors we are looking for can

have a common factor > 1. This eliminates lots of possibilities. For example: 2 4 or 2 6

or 2 2 or 2 12 or 3 6 or 3 12 or 6 2 or 6 3 or 6 4 or 6 12 (ignoring

signs temporarily). That's 10 of the 12 possibilities that can be eliminated without even multiplying

the two factors out. All we have left is (1 12 with 6 1) or (2 3 with 3 4).

Then since the rightmost term -12 in the product is negative we must make one of the numbers in

the rightmost column negative. (If the leftmost column of the answer is positive (here 6) then it is

customary to use only positive numbers in the leftmost column although we could make them both

negative.)

The 1 12 with 6 1 would produce a middle term of 71 or -71 (considering the signs) and

the 2 3 with 3 4 would produce a middle term of 1 or -1. The -1 is what we need.

So (2 -3)*(3 4) produces the 6 -1 -12 as desired; that is, (2x-3)(3x+4)=6x^2 - x - 12.

My high school algebra book (from the 1940's) taught factoring trinomials of this type this way

instead of the FOIL setup. I like it better and find that most of my students I show it to like it better also.

But it's a "take it or leave it" proposition. Choose the one you like best.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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