Without the use of the solve function of a calculator how do I solve the equation: x^4-5x^3-x^2+11x-30=0x45x3x2+11x30=0?

Given that (x-5)(x5) and (x+2)(x+2) are factors of f(x)f(x)

2 Answers
May 14, 2016

The zeros are x=5x=5, x=-2x=2, x=1+-sqrt(2)ix=1±2i

Explanation:

f(x) = x^4-5x^3-x^2+11x-30f(x)=x45x3x2+11x30

We are told that (x-5)(x5) is a factor, so separate it out:

x^4-5x^3-x^2+11x-30 = (x-5)(x^3-x+6)x45x3x2+11x30=(x5)(x3x+6)

We are told that (x+2)(x+2) is also a factor, so separate that out:

x^3-x+6 = (x+2)(x^2-2x+3)x3x+6=(x+2)(x22x+3)

The discriminant of the remaining quadratic factor is negative, but we can still use the quadratic formula to find the Complex roots:

x^2-2x+3x22x+3 is in the form ax^2+bx+cax2+bx+c with a=1a=1, b=-2b=2 and c=3c=3.

The roots are given by the quadratic formula:

x = (-b+-sqrt(b^2-4ac))/(2a)x=b±b24ac2a

= (2+-sqrt((-2)^2-(4*1*3)))/(2*1)=2±(2)2(413)21

= (2+-sqrt(4-12))/2=2±4122

= (2+-sqrt(-8))/2=2±82

= (2+-sqrt(8)i)/2=2±8i2

= (2+-2sqrt(2)i)/2=2±22i2

=1+-sqrt(2)i=1±2i

May 15, 2016

Let us try without knowing that (x-5)(x5) and (x+2)(x+2) are factors.
The constant term equals the roots product,so
30 = r_1*r_2*r_3*r_430=r1r2r3r4.

This coefficient is an integer value whose factors are pm 1, pm 2,pm 5, pm3±1,±2,±5,±3 Trying those values we can see that
p(-2) = p(5) = 0p(2)=p(5)=0 obtaining two roots.
We can represent the polynomial as
x^4 - 5 x^3 - x^2 + 11 x - 30=(x-5)(x+2)(x² + a x + b)

Calculating the right side and comparing both sides we obtain
-5=a-3
-1=b-3a-10
11=-10a-3b
-30=-10b

Solving for (a,b) we get a=-2,b=3
Evaluating the roots of x^2-2x+3=0 we get 1 - i sqrt[2],1 + i sqrt[2]