Find all solutions, both real and complex, to the following equation.

Question

My homeowrk says to use the quadratic equation to help solve, but I don'w know how to. Thanks in advance!

 

x^4 + x^3 - 5x^2 - 17x - 12 = 0

Answer 1

Brianna, I like to use synthetic division to determine possible factors of the polynomial. The possible roots can be determined by the factors of the last term being 12 in which are 1,2,3,4,6,12, and the factors of the coefficient of the first term being 1. Lets guess (x-3) is a factor of the polynomial. Use synthetic division.

 

  3    1   1   -5   -17   -12

            3   12    21     12

        1  4     7      4     0   Hence, x-3 is a factor of the polynomial in which could be rewritten as:

 

(x-3)(x^3+4x^2+7x+4)=0   x=3 is one of the solutions

 

Repeat the same process for the resulting 3rd degree polynomial. Since all the coefficients are positive, then a negative number would be a good guess. Let's guess x=-1 in which the possible factor being (x+1)

 

 -1   1   4   7   4

           -1 -3  -4

       1   3   4   0   Hence, x+1 is a factor in which the polynomial can be rewritten in its factored formed.

 

(x-3)(x+1)(x^2+3x+4)=0  x=-1 is another solution. Now you could use the quadratic formula to determine the rest of the roots.

 

x=(-3±√(3^2-4(1)(4)))/2 =(-3±√(9-16))/2=(-3±√-7)/2

 

Thus the last two two roots are complex numbers is which are:

 

x=-3/2 + i*1/2*√7  and x=-3/2 - i*1/2*√7

 

Hope this helps.

       

Answer 2

Hello Brianna -

 

This is a case where rather than just solving the problem, you have to use a certain method. For practice, that can be useful, though the point is to get the answer by whatever method works. In this case, the roots are -1, 3, and -3/2 +- i*sqrt(7)/2. Now let's use the QE to get them.

 

Fourth order equations can be solved by the quartic equation but it is quite long. To use the quadratic equation there are several options which involve variable substitution. In this case, since the leading coefficient is 1 (the x^4 term), it is a little simpler.

 

But in this case, as sometimes happens in practice problems, either 1 or -1 is a root. So by doing polynomial division (ask me if you need help on that), and trying division by (x + 1), we get x^3 - 5x -12. That in turn can be divided by (x + 3) to get x^2 +3x +4, which easily goes into the QE.

 

If you were to use a graphing calculator to graph the original equation, you would see it crosses the x-axis at -1 and 3. These are the real roots and imply factors of (x +1) and (x -3).

 

So I hope I helped. Let me know if you need more explanation.

 

Ken