complex zeros and factored form

Question

x³ - 5x² +17x-13 find the complex zeros and put it into factored form

Answer 1

The strategy here is to find the sole real zero, use the zero with division to reduce the polynomial to degree two.  The complex zeroes, if any, will be found by using the quadratic formula on the second degree polynomial.

 

1.  If there are real, rational zeroes, they are factors of 13.  -1, +1, 13, or -13.  Substituting 1 gives 1³ -5*1² + 17*1 -13 = 0.

 

Dividing x-1 into x³ - 5x² +17x - 13 (using either long division on synthetic division gives  x² - 4x +13;

 

Applying the quadratic formula gives  (4 +- √16 - 4(1)(13) )/2

 

   (2 +-3i)  are the two complex roots

Answer 2

Sophia,

 

You can first try factor (x - 1) and do long division on your expression to get (x - 1)(x² -4x + 13) = 0.

Now solve this for the 3 roots x₁, x₂, x₃ and put them into the fully factored form (x - x₁)(x - x₂)(x - x₃) = 0.

 

By the partial factorization in the first line, x₁ = +1, is the first zero and a real root.

 

Now use the quadratic formula to solve for x₂ & x₃ in the reduced equation (x² -4x + 13) = 0

So  x₁, x₂ = {+4 +- √[(-4)² - 4(1)(13)]}/2 = +2  +- (√-36)/2  = (2 +- 3i) , 2 imaginary zeros.

 

Finally [x - 1][x - (2 + 3i)][x - (2 - 3i)] = 0