Algerbra

Nevermind, you are factoring not solving.

If you don't know I can tell you that if the last term is positive it means that both factors will have the same sign (see how they are both negative). And the sign of the second term shows you what sign the factors will have.

And if the third term is negative it means you will have a positive and a negative factor.

PurpleMath has a really good lesson on factoring quads: http://www.purplemath.com/modules/factquad.htm

15t^2-67t+38

factors into

(5t-19)(3t-2)

I. Look at t^2 first combos (usually on a test it will be 1 x t^2, so this step will be eliminated)

1t, 15t

-1t, -15t,

3t, 5t,

-3t, -5t.

II. Then look at integer or number combos,

1, 38,

-1, -38,

2, 19,

-2, -19

III. So with your subsets for the t^2 and integers formulated, just plug in combos to find your cross multiplied t x numbers.

a. by logic, positive t^2 and positive integers with negative t cross multiplied would mean one is both negative and one is both positive (+ * - = -), so that reduces your combos in half.

1t, 15t, -1, -38 (t-1)(15t-38)-> -53t, nope, (t-38)(15t-1)-> -571t, nope

...........-2, -19 (t-2)(15t-19)-> -53t, nope, (t-19)(15t-2)-> -287t, nope

-1t, -15t, 1, 38 (1-t)(38-15t)-> -53t (skip, same as above)

............2, 19

3t, 5t, -1, -38 (3t-1)(5t-38)-> -119t, nope, (3t-38)(5t-1)-> --193t, nope

............-2, -19 (3t-2)(5t-19)-> -67t, BINGO , (skip remainder this subset)

(just for funzies, try the reverse)

-3t, -5t. 2, 19 (2-3t)(19-5t)-> -67, BINGO AGAIN for a second correct answer.

The problem is compounded if teacher pricky enough to make you do iterations with variable other than 1xt^2 on a test. OK for homework, you have time to work through. Tough on one-minute-per-problem test though, ha-ha.

This is off-the-cuff and there may be a more direct way to reduce subset. Of course writing will be much faster than typing too. Just practice all you can and learn common combos to try first to hit on first or second try.

algebrahelp.com is incredible... it helped me get to college level math after being out of school for 8 yrs...you can use their calculator to figure things out if you get stuck. There's also lessons and quizzes...good to use as a reinforcement of what you're learning in class.

I'm taking an Intermediate Algebra course right now, and the instructor taught us a method for factoring when you have a coefficient in front of the squared term that I had never learned before. I just scored a 100 on our test over factoring, and I contribute it to that method, because without it I would have run out of time doing trial and error.

It's called the A*C method, and I'll use your problem as an example-

1. First step is to make sure it's in ax^2+bx+c form (where a, b, and c are integers), this one already is.

2. Next, factor out GCF, in this problem it's one, so again, already done.

3. Multiply a and c.

15 x 38= 570

4. Find factors of this product that add up to b. I usually start with one unless it's a large number, such as this one, in which case I start with 5.

570= 5 x 114

6 x 95

10 x 57

I stopped here because when both are neg,

they add to -67.

5. Re-write problem in 4 terms using those factors.

15t^2-10t-57t+38

6. Factor by grouping and GCF. In this case it works to use it in this order, but sometimes it's easier to rearrange the two middle terms.

5t(3t-2)-19(3t-2)

7. Now factor out the GCF again, which is (3t-2), and you have your answer.

(5t-19)(3t-2)

Hope that helps! If you need any clarification, just let me know.

Laura

Great. What about the answer of (2-3t)(19-5t)? I could see that doing the iterative method (not trial and error). It is just a rule I suppose of the double negatives, but something to be aware of (even though these problems will never come up in real life).

Great method though. Thanks.

I understand what you're saying, but on this level of math, the main point of learning factoring is so that you can solve a quadratic equation. Using the answer given OR the second answer that you came up with, the values for t are the same.

On another note, if any of us end up needing to take a higher level of math, then couldn't you just use the A*C method, and then multiply each factor by -1? That would then give the same factors that you gave as a second answer. My curiosity is piqued, but only because I may take the higher level maths for the heck of it. That all depends on time and money, but I actually like math because it can be such a challenge, keeps the gray matter healthy.

Laura

"the values for t are the same."

BINGO, Thanks again.

Thank you too! I may have eventually realized that there is a second possible set of factors myself, but I would have to have time to sit down and really think about it, which is not very likely. The class I'm taking now is really fast paced, so I'm thankful that I have a head for math. Even so, it's been nearly 15 years since I started learning some of this stuff (I started in 8th grade, and I graduated h.s. 10 years ago), so really it's like learning it all over again, and I'm working for my A.

My instructor may not thank you though, because now you've thrown more questions into my head:D. I just e-mailed her with a link to this thread. I'm really curious if this is something that I may need to know in a future class, and if the A*C method could still be used. Way back in 8th grade, I learned the method that you showed, which to me is still trial and error, just "logical" and/or "educated" trial and error. I like short cuts, because I take forever just to double and triple check my work, so I need the extra time.

Thanks again, I always love it when someone makes me use my brain.

Laura