F(x)=x4-7x3+13x2+x-20 answer this pleaseI

Question

Find the zeros and roots

Answer 1

The first thing to note about the coefficients is that we have 1, 7, 13 and 1, 20, and 1+7+13=1+20. This suggests that plugging in x=1 or maybe x=-1 might make F(x)=0.

As it happens, F(-1)=1+7+13-1-20=0, so x+1 is a factor of the polynomial, making x=-1 a zero. We can use synthetic division to reduce the quartic to a cubic:

-1 | 1 -7 13    1  -20

      1 -1   8 -21 | 20

      1 -8 21 -20 | 0 = x³-8x²+21x-20. Let f(x)=x³-8x²+21x-20.

It's possible that we have some rational zeroes here. To find them let's look at the number 20 which has factors 2, 4, 5, 10 as well as 1 and 20. f(1)≠0, so let's look at f(2) and f(-2): f(2)=8-32+42-20=-2 and f(-2)=-8-32-42-20=-102, so neither 2 or -2 is a zero.

Let's try x=4: f(4)=64-128+84-20=0, so x-4 is a factor, and 4 is a zero so we can reduce the cubic to a quadratic:

4 | 1 -8  21  -20

     1  4 -16 | 20

     1 -4    5 |  0 = x²-4x+5.

We can easily reduce the quadratic to its factors by solving for x:

x²-4x+5=0,

x²-4x=-5, completing the LHS square:

x²-4x+4=-5+4

(x-2)²=-1.

Because -1 is negative we are in the world of complex numbers. Take the square root of each side:

x-2=±i where i is the imaginary √-1, Therefore x=2±i.

We now have the four zeroes: -1, 4, 2+i, 2-i.

F(x)=(x+1)(x-4)(x-2-i)(x-2+i)

We can check this by expanding the factors: (x²-3x-4)(x²-4x+5)=

x⁴-4x³+  5x²

   -3x³+12x²-15x

          -  4x²+16x-20 = x⁴-7x³+13x²+x-20=F(x), which confirms the discovered zeroes.