find all zeros for the function: f(x) =x^4-4x^3+x^2+16x-20

Question

(x) =x⁴-4x³+x²+16x-20

Answer 1

Because it is a polynomial to the 4th power you will have 4 answers. The possible roots of 20 are +/-1, +/-2, +/-4, +/-5, +/-10 and +/-20. Let's start with 2. Using long division of polynomials we get:

 

                (x⁴-4x³+x²+16x-20)/(x-2)=x³-2x²-3x+10  with no remainder, so x=2 is a root.

Now let's use x=-2:

 

                (x³-2x²-3x+10)/(x+2)=x²-4x+5 with no remainder, so x=-2 is a root.

We have 2 answers and we need 2 more. Because x²-4x+5 cannot be factored out we can use the Quadratic Formula to find the remaining answers. Remember, the Quad. Formula is: 

                                                  ax²+bx+c=0

 

                                                  x=(-b±(sq root of b²-4ac))/(2a)

                                                  x=(-(-4)±(sq root of (-4)²-(4)(1)(5)))/((2)(1))

                                                  x=(4±(sq root of (16-20))/2

                                                  x=(4±(sq root of (-4))/2

                                                  x=(4±2i)/2                    *i = sq. root of -1

                                                  x=2±i

Therefore, your roots are ±2 and 2±i.

 

Answer 2

The rational roots of f(x) must divide 20.

Try 2: f(2) = 16 -32 + 4 + 32 -20 = 0, i.e. 2 is a root.

Divide out the root to get: x³ - 2x² - 3x - 10

The rational roots of this equation must divide 10.

Try -2: - 8 - 8 + 6 +  10, i.e. -2 is a root.

Divide out the root to get: x² - 4x + 5 which by the quadratic formula gives the two roots as 2 ± 2i