Expert: Michael Tadesse - 9/13/2009

Question
QUESTION: I have some math questions that I attempted to do but I am pretty sure I did wrong for homework.
Could you please show me how to do them and give me the answers?

1) 5t²+15t+10
2) 18x+12x²+2x³
3) 2x-2xy²
4) 3t³-27t
5) 24a²-30a+9

An example question I did from this worksheet for how we are supposed to factor it is:

2x²+12x-80
=2(x²+6x-40)
2 numbers to multiply to get -40 and add to get 6:10 and -4
=2(+10)(x-4)

Thanks a lot!



ANSWER: Hello Kaitlyn

Factoring out polynomials is very important; it helps you to find their solutions.

Problem #1
5t2 + 15t + 10
5(t2 + 3t + 2) factor out 5
Search for two numbers that give 2 when multiplied and 3 when summed.
They are 1 and 2.
5(t +1) (t+2)


Problem #2
18x + 12x2 + 2x3 -- You see that x is common for the three algebraic expressions
2x (9 + 6x + x2) --- Now workout the included polynomial
2x (x2 + 6x + 9) --- Think of two numbers which 9 is their product and 6 is their sum, it is    only 3 Right?
2x (x +3) (x +3) = 2x (x + 3)2

Problem #3

2x – 2xy2 --- Factor out 2x
2x (1 – y2) – We can’t go further.

Problem #4

3t³ - 27t --- We can only factor out 3t (just look for something present everywhere in the algebraic expression, common to all expressions indeed.)

3t (t2 – 9) ---- Kaity, can’t we factor out the polynomial inside the bracket? What do you think?

Think of two numbers that give -9 when multiplied and 0 when added.
The middle term is 0 that is why I said their sum is 0.
They are 3 and its additive inverse, -3 aren’t they?
Therefore,

3t (t + 3) (t – 3), when factored out explicitly.

ARE MY STEPS CLEAR KAITILYN? IF NOT LET ME KNOW AND TRY AGAIN





---------- FOLLOW-UP ----------

QUESTION: Thanks a lot for the help with the first four.
Three of them that I tried I actually did right, but I just didn't think they seemed right!
Could you please help me with number five though?

Answer
Hello again Kaitlyn

Sure I can help you with number five.


24a²-30a+9

Kaitlyn, before I get on with the problem I have to recall you the quadratic equation formula.

Sometimes it is not possible or readily apparent how to factor an equation or complete its square. However, all quadratic equations that can be put into the form
         ax2 + bx + c = 0
Can be solved using the quadratic formula:
There is an image representing the quadratic formula download it and use it.
So let’s continue
First factor out 24
24 (a2 – 5/4 a + 3/8) isn’t it? Just divide every term by 24
Now let’s deal with the inner polynomial, we have to factor it out.
Here it is.
a2 – 5/4 a + 3/8 Now think of two numbers that 3/8 is their product and 5/4 is their sum.
It’s a bit baffling isn’t it?
So here is what you do. You have to use that image you just downloaded (which represents the quadratic formula) to find the solution of the polynomial. Then we can factor the polynomial with the solutions at hand.
The formula is designed for X, but we can use it for a as the variable.
Here, a = 1(the coefficient)
        b = -5/4
        c = 3/8
So substitute them in the formula and you’ll find two answers, {½, and ¾}
Therefore the inner polynomial can be written as
(a - ½) (a – ¾) = a2 – 5/4a + 3/8
Now let’s not forget 24.
24 (a - ½) (a – ¾)
This is the way you factor it. Try to multiply this and check the exactness.
I know it is too long but try to observe each step carefully.
Finally, I started to give some tutorials (questions with answers and short notes) to some students. I preferred the postal system because most of the work is done on paper. If you want to be one of them just send me a notification email.
millimike@technologist.com
         ALWAYS READY TO HELP YOU