(1) x5-3x4-24x3+72x2-25x+75=0 (note corrections).
First look for rational zeroes. 75=3×5×5, so rational zeroes could be 3 or 5 (positive or negative forms).
Try x=3: 243-243-648+648-75+75=0, so x=3 is a zero. Divide by this zero:
3 | 1 -3 -24 72 -25 75
1 3 0 -72 0 | -75
1 0 -24 0 -25 | 0 = x4-24x2-25 = (x2-25)(x2+1)=(x-5)(x+5)(x2+1).
The expression has 3 real zeroes: 3, 5, -5 and two complex zeroes: i, -i.
Another way is to spot that the expression can be written by grouping pairs of terms:
x4(x-3)-24x2(x-3)-25(x-3)=(x-3)(x4-24x2-25), and then continue as above.
(2) 5x5-20x4-4x3+16x2-45x+180=0 (note suggested corrections),
5x4(x-4)-4x2(x-4)-45(x-4)=0=(5x4-4x2-45)(x-4). x=4 is a real zero.
5x4-4x2=45,
5(x4-⅘x2)=45,
x4-⅘x2=9,
x4-⅘x2+4/25=9+4/25 (completing the square),
(x2-⅖)2=229/25, x2-⅖=±√229/5, x2=(2±√229)/5, x2=3.426549 (negative zero has no real square root). Therefore, x=1.8511 or -1.8511 (approx). Real zeroes are 4, 1.8511 or -1.8511.
(3) -3x^5+3x^4+7x^3-7x^2+12x-12=0 (suggested correction),
-3x4(x-1)+7x2(x-1)+12(x-1)=0=(-3x4+7x2+12)(x-1). x=1 is a zero.
-3x4+7x2+12=0, so 3x4-7x2-12=0. Using the quadratic formula:
x2=(7±√(49+144))/6=(7±13.892444)/6=3.482074 approx, so x=±1.8660.
Real zeroes are 1, 1.8660, -1.8660.
(4) x3+x2=15x-15, x3+x2-15x+15=0. Shouldn't this be x3+x2=15x+15 or x3+x2=-15x-15? As it stands a completely different solution method would be required.
Note that x3+x2=15x+15 becomes x2(x+1)-15(x+1)=(x2-15)(x+1)=0 with zeroes: ±√15 and -1. x3+x2=-15x-15 becomes x3+x2+15x+15=(x2+15)(x+1) with only one real zero: -1.
(5) x4-6x2-7x-1=0 has two real zeroes. If this equation has been correctly presented, solutions can be found using iterative methods. Approx solutions are -0.167 and 2.918.
Here is Newton's method for solving this as it stands. We need the differential=4x3-12x-7.
xn+1=xn-(xn4-6xn2-7xn-1)/(4xn3-12xn-7).
For the negative solution let x0=0.
x1=-(-1/(-7))=-1/7,
x2=-0.1658..., x3=-0.1665..., x4=-0.1665129426..., x5=-0.166512942598 (approx).
For the positive solution let x0=3. After several iterations we get x=2.918240841974.
If all the original equations were in fact correctly stated, the iterative method can be applied to each to find all the real zeroes. The first step is to sketch a graph of each, setting y equal to the expression. The zeroes are the x-intercepts. The graphs will provide close approximations which can each be used as x0 for the iteration process.