can some one please tell me how to factor this polynomial?: x^5 - 15x^3 - 10x^2 + 60x + 72?

Answer 1

For those that aren't able to use a calculator, you need to use the integer roots theorem to find possible roots. You can then use synthetic division to try each root until you find one that fits. Continue this process until all roots have been found. See below for more details:

Answer 2

Factor x^5 -15x^3 - 10x^2 + 60x + 72
Use synthetic division to try different factors:
(x+2)(x^4 - 2x^3 -11x^2+12x +36) works.
So now use synthetic division on the 4th degree polynomial and you will find (x-3) is a factor, so now we have:
(x+2)(x-3)(x^3 + x^2 -8x -12) Now , I'm going to leave it up to you to find the factors, if any, of x^3+x^2 -8x -12. Hint: There has to be at least one more factor that is real because a fifth degree equation has to have 5 roots and we have found two. That leaves three of which only two can be imaginary so one has to be real. You might look for a double root at x=-2 or x=3. In other words, try dividing x^3 +x^2 -8x -12 by either (x+2) or (x-3).

Answer 3

It will be very difficult unless it has integer roots.

If it does have integer roots, putting that value in for x will make the polynomial zero.

For a start, let's ignore signs, and just write down what the absolute magnitudes of the terms are when x is 1, 2, 3, . . . We can worry later about which sign to use.

(1) 1, 15, 10, 60, 72, total = 158, no way can we split that into 79 and 79.

(2) 32, 120, 40, 120, 72. LOOK AT THAT, it's screaming for the two 120's to have opposite signs, and the 32 and 40 to have the opposite sign to the 72. Check out -2 and 2 to confirm.

(3) 243, 405, 90, 180, 72, total 990, half of 990 = 495, LOOK AT THAT, it's screaming for the 405 and 90 to have one sign, and the rest to have the other, so check out -3 and 3 to confirm.

That's how to get started. -2 makes the polynomial zero, so (x + 2) is a factor; and +3 makes it zero, so (x - 3) is a factor. Divide them out by synthetic division (which is a separate question) and start again with what's left.

Answer 4

I would graph the equation and see where the the line crosses the x-axis when you graph the line you see it crossing two times but the end of the little parabola is counted twice, then i personally would synthetically divide where it crosses the x- axis into the original equation like this:
3 (where it crosses) [ 1 0 -15 -10 60 72] these come from the original equation , the zero from x ^4 which there is none so you have to put zero in. When you synthetically divide the one will stay there but then you multiply it by 3 then you add it to the 0 doing this for each of the numbers so it would look like this:
3 [ 1 0 -15 -10 60 72]
3 9 -18 -84 -72
1 3 -6 -28 -24 0
thus you took out (x-3) from the original equation leaving x^4 +3x^3 - 6x^2 -28x -24. THen you synthetically divide with three agian because it techically crosses twice thus you get (x-3)(x-3)(x^3 + 6x^2 + 12x + 8), if you look at the graph again you see that it crosses -2 so you synthetically divide -2 into (x^3 + 6x^2 + 12x + 8), which gives you x^2 + 4x + 4 now to figure out the answer of that you must use the quadratic formula. (-b +or- (sq root (b^2-4ac)))/ 2a and by doing that you should get your final answer

Answer 5

f(x)=x^5-15x^3-10x^2+60x+72
on acursory examination we find f(-2)=0
so x+2 is a factor
synthetically dividing by x+2
1 0 -15 -10 60 72
0 -2 4 22 -24 -72
1 -2 -11 12 36 0
the quotient is x^4-2x^3-11x^2+12x+36
x-3 is a factor
again dividing synthetically
1 -2 -11 12 36
0 3 3 -24 -36
1 1 -8 -12 0
the quotient is x^3+x^2-8x-12
again x-2 is a factor
1 1 -8 -12
0 -2 2 12
1 -1 -6
the quotient is x^2-x-6
onfactoring this will give
(x-3)(x+2)
sothe factors will be
(x+2)(x-3)(x+2)(x-3)(x+2)

Answer 6

x^5 - 15x^3 - 10x^2 + 60x + 72 factors into

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

ROOTS OF POLYNOMIALS

OF DEGREE GREATER THAN 2

http://www.themathpage.com/aPreCalc/factor-theorem.htm

Answer 7

My graph shows a zero at -2 and 3 so divide (x+2) and (x-3) from this polynomial and see what zeros are left. There may be multiple zeros at some of these points.

Answer 8

You should begin to try with some integral number as root.So you will find x=-2 is a triple root.That leaves you a final factor of second degree .
so your polynomial =(x+2)^·3 * (x-3)^2

Answer 9

Graph it on a calculator