Factor Completely: x^4-15x^3+81x^2-185x+150

Question

I'm confused on how to do this question

Answer 1

By trial and error, 3 is a root.  Divide out synthetically by x - 3:

 

3⌋     1         -15       81       -185        150

                       3      -36         135       -150                                     1        -12     45          -50           0

So, the given polynomial is equivalent to:

 

(x - 3)(x³ - 12x² + 45x - 50)

 

Again by trial and error, 5 is a root of the second factor.  Divide out synthetically by x - 5:

 

5⌋ 1      -12      45      -50

                5     -35        50                                

    1        -7       10       0

 

So, the given polynomial factors as:

 

(x - 3)(x - 5)(x² - 7x + 10)

 

= (x - 3)(x - 5)(x - 5)(x - 2)

 

= (x - 2)(x - 3)(x - 5)²    

Answer 2

There's no quick way.  First use the Rational Zero theorem to come up with a list of all the possible zeros of this polynomial (i.e., the solutions to the equation x⁴-15x³+81x²-185x+150 = 0). 

 

To do this, first find all the numbers that evenly divide 150.  (You don't have to be concerned with the coefficient of the leading term, since it is 1.)  These are: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

 

Use each of these to form a potential binomial factor, such as x-1, x-2, x-3, x-5, etc.

 

Then start to divide the original polynomial by each of these potential factors until you find one that divides it evenly.  Then you've found one of the factors, and the quotient should be a 3rd degree polynomial.

 

Do the same thing again with the quotient, so that you get another binomial factor and a new quotient, which will be a 2nd degree polynomial.  That should be easy to factor in the usual way into two binomial factors. 

 

When you've done all this, the four binomials will be the factors of the original polynomial.

 

But if you reach a point where any of the quotients can't be factored, then that quotient itself will be one of the factors, and the binomials you've found will be the others.