Help needed!

Marisca · 2012-03-30 22:50:12

Hi, I can't seem to solve this problem;

Factor f(x)=6x^3+5x^2-17x-6 completely.

Bob · 2012-03-31 01:09:27

hi Marisca,

(x+2) is a factor. Is that enough for you?

I used the factor theorem. Do you know this?

Bob

John E. Franklin · 2012-03-31 01:19:46

i'm a little rusty on my algebra, but i'll give it a try.
Usually for these factoring problems I simply guess at a factor and divide it to see if it comes out without a remainder.
So we start with your 6x**3 + 5x**2 - 17x**1 - 6x**0.
Notice ** and ^ are the same thing, i think ibm used ** a long time ago and i liked it.
Also notice that the 6 at the end i made 6x^0, and that is the same thing because
x up to the power of zero is just equal to the number one or 1 or unity.
And one times 6 is 6, so that is the same thing.
Now on deciding what to divide by, usually your best bet is to choose either (x-1) or (x+1) to start,
and then divide and see if you get no remainder. No remainder is good, because then it is a clean divide,
and it means you found a factor plus your answer is another equation you can take and work on more.
See (x-1) times (x**2 + x + 1) or something like that might come up, just guessing, and then you
would have to try to figure out if the (x**2 + x + 1) can be made into two smaller factors like (2x - 2)
or (3x + 3) or something by dividing again. But if you get a remainder, then your original guess
must be wrong, so don't use it as a factor.
So now i'll see if i can remember how to divide 6x**3 + 5x**2 - 17x**1 - 6x**0 by (x-1).
Hmmm... Let me go do some reading on the internet about dividing polynomials and then I'll
come back in fifteen minutes...

anonimnystefy · 2012-03-31 01:29:02

Another way to get a rational factors is to look at the first and the last coefficient.

If p/q is the root of a polynomial then then q must be a divisor of the first coefficient and p must be a divisor of the last coefficient.

John E. Franklin · 2012-03-31 01:56:08

wow, bob bundy is smart, but even though he did it, i'll keep trying to learn, so here goes nothing.

I will demonstrate dividing by a guess of (x-1), i think that is called the divisor or denominator.
So x from the (x-1) goes into 6x**3, 6x**2 times, so basically 6x**3 divided by x is 6x**2 or 6x^2, same thing. (meme chose)
(meme chose is in french for same thing)
Now write down the 6x^2 above the long division shape like this:
_______________
)

Put the 6x^2 above that whole thing just like long division in 4th grade.

Now just like long division in 4th grade with just numbers, you multiply the top number you just wrote
by the number of the left of the )

So (x-1) is to the left of the )

So 6x^2 times (x-1) is 6x^3 - 6x^2. (This step I'll name step AA11 in case you want to talk about that step.)

So write the AA11 answer below the equation (polynomial) we are starting with inside the
_________________
)

              6x**2
          ________________________________
(x-1)  )   6x**3 + 5x**2 - 17x**1 - 6x**0
             
             6x**3  - 6x**2

This is just the very beginning of dividing, it is faster on paper than by typing like this ofcourse.

The next thing to do is to compare this to a long division problem from when you were ten years old with numbers.

          8
     _____________
6  )   49
       
        48

See how they are similar to each other, 8 times 6 is 48, then you would subtract the 48 from the 49 back in school.
So that's what you do with algebra too.

I'm getting a little tired, but if you want me continue with this idea, just let me know and call it the AA22 conversation.

Bob · 2012-03-31 03:55:01

hi John E. Franklin

Thanks for the ringing endorsement, you are too kind!

Here's what I did.

step one: enter the function into the MIF graph plotter at

http://www.mathsisfun.com/data/function … =-8&ymax=8

That way you get an idea what the 'roots' might be.

Two looked like fractions so I thought I'd try x = -2

The factor theorem works like this.

If f(x) = (x-a) . g(x) then setting x = a gives f(a) = 0, so you can quickly test a possible factor.

In this case try x = -2

f(-2) = 6 x (-2)^3 + 5 x (-2)^2 - 17 x (-2) - 6 = 6 x -8 + 5 x 4 + 34 - 6 = -48 + 20 +34 - 6 = 54 - 54 = 0

So that confirms the suspicion that x = -2 is a root => (x + 2) is a factor.

Now you can use algebraic division to get the quadratic that goes with this; but, as you have seen, it is messy to write out.

As you know it's a quadratic, you can just manipulate the terms like this

(x+2) times (what?) = 6x^3 etc

Well it must start (6x^2 + ...)

Now +2 times 6x^2 = 12 x^2 but we want only 5x^2 so we need to 'lose' 7x^2. This means that the quadratic must continue like this

(6x^2 - 7x + ....)

But +2 times -7x = -14x and we want -17x so we need another -3x. So the quadratic continues

(6x^2 - 7x -3). If I've made no mistakes (useful check here) the +2 times -3 should give the constant term -6, which it does.

So f(x) = (x+2)(6x^2 - 7x - 3)

Now to factorise the quadratic. (check b^2 - 4ac = 49 + 4 x 3 x 6 = 49 + 72 = 121. This is a perfect square so the quad should factorise nicely without needing the full quad formula)

(x+2)(3x+1)(2x-3)

Looking back to the graph plotter, that looks about right, so probably no mistakes.

Bob

John E. Franklin · 2012-03-31 07:12:39

Nice, I tried out the function-grapher and the cubic seems to cross at about -2, -1/3, and +1 1/2.
Sure enough plugging those into bobbundy's three factors does in fact make each factor zero.
I forgot about that technique from years ago, but it is sweet to see it again, thanks!!

John E. Franklin · 2012-03-31 07:17:34

Oh, I am so excited by this idea!!
take the factors and remove the multipliers out and off of the x, and put the multiplier on the outside of the whole system.
wammo:
(x+2)(3x+1)(2x-3)
becomes
(x+2)3(x + 1/3)2(x-1 1/2)
so 2 times 3 is 6 so you get:
6(x+2)(x+1/3)(x-1-1/2) and then you can see the zeros right away without dividing!!
Thanks bobbundy!!

Bob · 2012-03-31 07:18:23

Looking at the graph has to be a guide only.

For example, the root might not be x = -2, but rather x = -1.99999997

A graph on its own is not definitive, because it does not have 'perfect' accuracy, but it is a good place to start.

Bob

Marisca · 2012-03-31 11:35:26

ah, thanks everyone!!
I definitely understand it now.

Math Is Fun Forum

#1 2012-03-30 22:50:12

Help needed!

#2 2012-03-31 01:09:27

Re: Help needed!

#3 2012-03-31 01:19:46

Re: Help needed!

#4 2012-03-31 01:29:02

Re: Help needed!

#5 2012-03-31 01:56:08

Re: Help needed!

#6 2012-03-31 03:55:01

Re: Help needed!

#7 2012-03-31 07:12:39

Re: Help needed!

#8 2012-03-31 07:17:34

Re: Help needed!

#9 2012-03-31 07:18:23

Re: Help needed!

#10 2012-03-31 11:35:26

Re: Help needed!

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