synthetic division?

Question

Can someone help me figure these 2 out.
1)x^7+16x^6+113x^5+450x^4+1080x^3+1552x^2+1232x+416

2)x^6-x^5-11x^4+13x^3+26x^2-20x-24

I know how to set these up but if you can tell me the factored answer and what you put in the upper left hand corner box i think i will have it. to figure out what goes in the upper left hand corner box you have to find all the factors of either -24 or 416. i have tried and tried and i just cant get it. PLEASE HELP ME!

Answer 1

Oi. Please put spaces in your polynomials so Yahoo doesn't cut them off.

Now, we have:

#1: x^7 + 16x^6 + 113x^5 + 450x^4 + 1080x^3 + 1552x^2 + 1232x + 416

Now, this is a nasty polynomial, but let's make some observations. First, all the coefficients are positive, so if x is any positive number, then P(x) is positive (where P is the above polynomial), so x cannot possibly be a root of P. This means we only have to consider negative roots. First, let's find the factors of 416:

416
208*2
104*2^2
52*2^3
26*2^4
13*2^5

Okay, so the prime factorization of 416 is 13*2^5, so the possible factors of 416 we have to test are -1, -2, -4, -8, -16, -32, -13, -26, -52, -104, -208, and -416 (we only have to test the negative factors because of the observation above). All I can say is, this is going to get extremely tedious if it doesn't turn out to be one of the first three, which means that one of the roots must be either -1, -2, or -4, because anything else would result in the student doing an unreasonable amount of work (this is some metamathematical reasoning that you should gain more experience with as you progress further).

So let's try -1:

-1| 1 16 113 450 1080 1552 1232 416
......... -1 -15 ..-98 .-352 .-728 -824 -408
--------- --------- ----------- ---------- ---------- -
..... 1 15 ..98 352.. 728 ..824 .408 .....8

Well, we got a remainder, so -1 was not a root. Let's try -2:

-2| 1 16 113 450 1080 .1552 .1232 ..416
........ -2 .-28 -170 -560 -1040 -1024 -416
---------- --------- --------- --------- --------- ------
..... 1 14 ..85 280 ..520 ...512 ...208 .......0

All right! -2 is a root, which means we've got a factor of x+2. Of course, now we've got a smaller (but still pretty large) polynomial to factor. We know -1 isn't a root, but -2 still might be, so let's try it again:

-2| 1 14 .85 ..280 520 ..512 .208
........ -2 -24 -122 -316 -408 -208
------------- ----------------- -------------
.....1 12 .61 ..158 204 ..104 ......0

Another factor of x+2 out. And trying -2 as a root again:

-2| 1 12 61 158 204 104
........ -2 -20 -82 152 -104
-------------- ------------ --------
.....1..10 41 ..76 ..52.... 0

Another factor of x+2 out. And trying -2 as a root again:

-2| 1 10 ..41 .76 .52
......... -2 -16 -50 -52
----------- ---------------
.....1 ...8 ..25 26 ...0

A fourth factor of x+2 out. And trying -2 as a root again:

-2| 1 8 .25 .26
...... -2 -12 -26
--------------------
.... 1 6 .13 ....0

So a fifth factor of x+2 out. Of course, -2 does not divide 13, so we have no more factors of x+2, but no matter. At this point our remaining polynomial is the quadratic equation x²+6x+13, so using the quadratic formula:

x = (-6±√(36 - 52))/2
x = (-6±√(-16))/2
x = (-6±4i)/2
x = -3±2i

So x²+6x+13 factors as (x+3-2i)*(x+3+2i). Thus the entire polynomial can be written as:

(x+2)^5 * (x+3-2i) * (x+3+2i)

#2: This one is just as hard as the first one. First, we factor 24:

24
12*2
6*2^2
3*2^3

So possible roots we need to try are ±1, ±2, ±4, ±8, ±3, ±6, ±12, and ±24. Now, again, we know that some of the first few roots above have to be factors, because otherwise this would result in an unreasonable amount of work. We start with 1:

1| 1 -1 -11 .13 26 -20 -24
....... 1.....0 -11 ..2 .28 ....8
--------------- -------------------
... 1.. 0 -11 ..2 28 ...8 -16

Nope. Let's try -1:

-1| 1 -1 -11 13 .26 -20 -24
...... -1 ....2 ..9 -22 ..-4 ..24
------------ ------------ -------------
.....1 -2 ..-9 22 ....4 -24 ...0

Alright. A factor of x+1 extracted. Trying again:

-1| 1 -2 -9 22 ...4 -24
....... -1 3 ..6 -28 ..24
-------------- ---------------
.... 1 -3 -6 28 -24 ...0

Alright. Another factor of x+1 extracted. Trying again:

-1| 1 -3 -6 28 -24
...... -1 .4 ...2 .30
-------------- ------------
.... 1 -4 -2 30 ...-6

Okay, we get a remainder, so -1 is not a factor. We already tried 1, so the next logical factor is 2:

2| 1 -3 -6 ..28 -24
....... 2 .-2 -16 .24
------------ ------------
.... 1 -1 -8 12 .....0

Alright. So x-2 is a factor. Trying again with 2:

2|1 -1 -8 12
..... 2 ..2 -12
-----------------
... 1 1 -6 ...0

So x-2 is a factor. At this point the remaining polynomial is x²+x-6, which factors easily as (x-2)(x+3). So the final factorization is:

(x+1)² (x-2)³ (x+3).

And we are done.

Answer 2

2) x^6 - x^5 - 11x^4 + 13x^3 + 26x^2 - 20x - 24

Since the leading factor is one, the factors all have to multiply to 24. If the factors are rational, and they are, the only possibilities are

±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24,

Since we need six factors that multiply to 24, I would start with the lower ones, they are more likely. The expression factors as:

(x + 1)²(x - 2)³(x + 3)

The first one works the same way.