x^7 + 2x^6 + 7x^5 + 14x^4 - 17x^3 - 34x^2 + 9x + 18 = 0?

Question

find all real and imaginary roots of the polynomial below using any techniques. (hint:this problem is very repetitive) state the multiplicity of each root.

x^7 + 2x^6 + 7x^5 + 14x^4 - 17x^3 - 34x^2 + 9x + 18 = 0

write answer in simplest form. show all work.

Answer 1

becuase ratio of successive terms are 1: 2 so we get
= x^6(x+2) +7x^4(x+2)-17x^2(x+2)+9(x+2)
= (x+2)(x^6+7x^4-17x^2+9)
now x^6+7x^4-17x^2+9 is cubic in x^2 say w

w^3+7w^2-17w+9

by putting w = 1 we get f(w) = 0

so (w-1) is a factor

w^3+7w^2-17w + 9
= w^3-w^2 + 8w^2-8w - 9w + 9
= w^2(w-1)+8w(w-1)- 9(w-1)
= (w-1)(w^2+8w-9)
= (w-1)(w+9)(w-1)
= (w-1)^2(w+9)

so original function = (x+2)(x^2-1)^2(x^2+9)
= (x+2)(x-1)^2(x+1)^2(x+3i)(x-3i)

roots are -2 , +1 (double) -1 (double) ,-3i, + 3i

we have found all 7 roots

Answer 2

First, let us list the possible rational roots of this polynomial:

±1, ±2, ±3, ±6, ±9, and ±18

Now let us test them by synthetic division, beginning with the simplest:

1| 1 2 7 ..14 -17 -34 ...9 18
...... 1 3 ..10 24 .... 7 -27 -18
--------------------------------------
... 1 3 10 24 ...7 -27 -18 0

So 1 is a root of the polynomial. One root down, six to go. Since the problem hints that it is very repetitive, perhaps we should check to see whether this root repeats itself. We do this by taking the quotient of our last division and seeing whether (x-1) divides that:

1| 1 3 10 24 . 7 -27 -18
...... 1 . 4 14 38 45 ..18
--------------------------------
... 1 4 14 38 45 18 0

Well, it seems 1 was indeed a root of multiplicity 2. Of course, that's all of them, our quotient polynomial's coefficients are all positive, so it cannot possibly have any positive roots. Let us start testing the negatives. Let's start with -1:

-1| 1 4 14 38 ..45 18
....... -1 -3 -11 -27 -18
----------------------------
.....1 3 11 27 ..18 0

Lo and behold, -1 is a root. 3 down, 4 to go. Let us see if -1 repeats:

-1| 1 3 11 27 18
....... -1 -2 .-9 -18
-----------------------
.... 1 2 ..9 .18 0

And so -1 repeats itself as well. Four down, 3 to go. Unfortunately, -1 does not repeat itself a third time. However, we still have other possible rational roots. Let us see if one of them will work, starting with -2 (since it is the simplest of the remaining possibilities):

-2| 1 2 9 18
.......-2 0 -18
------------------
.....1 0 9 0

All right, -2 works out to be a root as well. And at this point we are left with a quadratic equation x²+9=0, which may easily be solved to reveal that x=±3i. Therefore, the complete set of roots, listed with multiplicities, is:

x={1, 1, -1, -1, -2, 3i, -3i}

And we are done.

Answer 3

I'd look for rational roots first...
Rational Zero Theorem says any rational root must be in the form p/q where p is a factor of 18 and q is a factor of 1.
This leaves ± {1, 2, 3, 6, 9, 18} as possible rational roots.
Plugging these in (merci Excel) we see that -1, 1, and -2 are roots.
After dividing by (x-1) I found that the resulting polynomial also had 1 as a root - so our roots are now -1, 1, 1 and -2.

Dividing this monster by (x-1)²(x+1)(x+2) results in...
x³ + x² + 9x + 9 = 0
→ (x²+9)(x+1) = 0
whise roots are ±3i and (again) -1.

Answer: -1, -1, 1, 1, -2, 3i, -3i.

Answer 4

Rational root theorem suggests ±1,±2,±3, and ±6 as possible roots. I set up a simple Excel spreadsheet to do synthetic division and found ±1 and -2 to be rational roots. The depressed equation was then

x^4 + 8x² - 9 = 0
(x² + 9)(x² - 1) = 0,

which gives us another pair of x = ±1, and also x = ±3i.

so roots are 1(twice), -1(twice), -2, and 3i and -3i.

Answer 5

given x^2 + 9x + 18 = 0 you want to split this into 2 factors (x+ )(x+ ) = 0 what 2 numbers even as further grants you with 9 or perhaps as elevated grants you with 18? (x+3)(x+6) = 0 set each and anytime period equivalent to 0 x+3 = 0 and x+6 = 0 and clean up x = -3 and x = -6

Answer 6

By inspection, x=1...

Answer 7

Do it yourself