MATH HW Q:5 TO CHAPTER 6

Question

Find a polynomial with rational coefficients of minimal degree passing through (−2,7) and having zeros

1, and 1/2, and 1/2, and 1 − squareroot of 2i, and 2 + squareroot 3𝑖.

Answer 1

To add some detail and notation to Raymond's reply:

1. Assume that the double-mention of 1/2 is intentional, i.e. there is a double zero at x=1/2. Therefore the desired polynomial - say, p(x) - has a factor (2x-1)²
2. By the Complex Conjugate Root Theorem, since p(x) has real coefficients, if a+bi is a root, then so is a-bi . Thus, the two given complex zeroes yield four factors of p(x) : x-(1-√2 i), x-(1+√2 i), x-(2+√3 i), x-(2-√3 i). Multiplying these gives: (x²-2x+3)(x²-4x+7) .
3. Not forgetting about the zero at x=1, we know that p(x) has factors (x-1)(2x-1)²(x²-2x+3)(x²-4x+7) . Any constant real multiple of this expression will not change its zeros, so we have p(x) = c(x-1)(2x-1)²(x²-2x+3)(x²-4x+7) for some real c. Since the curve passes through (-2,7), we have p(-2)=7 . Substituting x=-2, get: 7 = c(-3)(-5)²(11)(19), therefore c = -7/15675.
4. In factored form, p(x) = (-7/15675)(x-1)(2x-1)²(x²-2x+3)(x²-4x+7) . Multiplying the terms of highest degree shows that the polynomial has degree 7. Expanding this is left as an exercise for the student :) .

Answer 2

it has 7 roots or zeros. 3 real, 4 imaginary

it's 7th degree

x=1-isqr2

x-1 =-isqr2

x^2-2x+1=-2

x^2-2x +3 is one factor

x^2+4x+7 is another factor

(x-1/2)^2 and

x-1 are 2 more factors

multiply all 4 factors together, with coefficient "a"

f(x)=a(x-1)(x-1/2)^2(x^2-4x+7)(x^2-2x+3) =7 when x=-2

solve for "a"

-3a(25/4)(19)(11)=7

a= -7/3(209)(25/4)

a =-28/(209)(75)

f(x)=-(7/15675)(x-1)(2x-1)^2(x^2-4x+7)(x^2-2x+3)

assuming "squareroot 2i and 3i" meant isqr2 and isqr3