Finding unknown values and solving Q(x) = x^3 - 6x^2 + ax + b

Question

Question:

The polynomial Q(x) = x^3 - 6x^2 + ax + b has one zero (1 - i√5), where a and b are real. Find the values of a and b and solve the equation for Q(x) = 0. Note that 'i' represents the complex number i.

 

I understand that if 1 - i√5 is a root then the conjugate (1 + i√5) must also be a root. I thought then that I maybe just needed to sub in Q(1 - i√5) and Q(1 + i√5) and make it equal to zero in order to find the values of a and b, but my answer was quite long and complicated.

Answer 1

The method below provides a way of solving the problem without the use of division.

 

The two complex roots are 1-i√5 and 1 + i√5

Sum of roots = 2

Product of roots = 6

 

Quadratic expression = x² - (sum Roots) + Product Roots = x² - 2x + 6

The required cubic polynomial is of the form

P(x) = (x² - 2x + 6) (Ax + B) = Ax³ + (B-2A) x² + (6A - 2B) x + 6B

 

Compare the coefficients of the x's in P(x) to the coefficients of

Q(x) = x³ - 6x² + ax + b

From these two equations, we get

A =1

B-2A = -6      ⇒ B - 2(1) = -6   ⇒  B = -4

a = 6 A - 2B = 6(1) -2(-4) = 14

b = 6B = 6(-4) = -24

Answer 2

The easiest way (although at times contrary to popular belief) is to do long divison to find the last factor which will be independent of a and b as the remainder is zero by definition. First determine your quadratic factor from imaginary roots:

 

Factor: (x - 1 - isqrt (5))(x - 1 + isqrt (5))

= x^2 - 2x + 6

 

x^3 -6x^2 + ax + b ÷ x^2 - 2x + 6

= x - 4 (Note that a and b just ensure that the remainder is zero)

 

Then, just expand the factors:

Q (x) = (x - 4)(x^2 - 2x +6)

  =  x^3 - 6x^2 + 14x - 24

Thus, a = 14; b= -24

 

And when Q (x) = 0

x = 4, 1 - isqrt (5), 1 + isqrt (5).

 

Answer 3

Hi Chloe,

 

Since the coefficients of Q(x) are real and one zero is (1-i√5), then another zero must be (1+i√5). Also since (1-i√5) and (1+i√5) are zeros, then (x- 1-i√5) and (x- 1+i√5) are factors of Q(x). Multiply these two factors to get x² -2x +5 which is a factor of Q(x). To find the remaining factor use long division of polynomials  (x³ -6x² +ax +b)÷(x² -2x +5)

 

                                 x___-___4_____

           x² -2x +5     |  x³ - 6x² + ax +b

                                x³ - 2x²  + 5x 

                               ------------------

                                    -4x² +(a-5)x +b

                                    -4x² + 8x    - 20   Since x² -2x +5 is a factor of Q(x) the remainder must be zero.

                                    --------------------

                                               0   +  0    Therefore,  a - 5 -8 = 0    and  b - (-20) = 0  and

                                                                                       a = 13                   b = -20