Find all real zeros of x^5-3x^4-24x^3-72x-25x+75

Question

could you please also answer these??  please I'm really desperate:

5x^5-20x^4-40x^3-16x^2-45x+180=0

-3x^5+3x^4+9x^3-7x^2+12x=12

x^3+x^2=15x-15

x^4-6x^2-7x-1=0

Answer 1

(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.

x_n+1=x_n-(x_n⁴-6x_n²-7x_n-1)/(4x_n³-12x_n-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.