Question

im2301_mvoa integrated math iii semester a_2439883
writing polynomial functions from complex roots
$(x + 1)(2 + ix)(2 - ix)$
$(x + 1)x + (2 + i)x - (2 - i)$
$(x - 1)x + (2 + i)x - (2 - i)$
$(x - 1)(2 + ix)(2 - ix)$
the polynomial function of least degree with a leading coefficient of 1 is:
$f(x) = x^3 + ax^2 + bx + c$
$a = \square$
$b = \square$
$c = \square$

Explanation:

Step1: Simplify the quadratic factor

First, we simplify the product of the complex factors. Let's take the correct polynomial form (assuming the roots are -1, \(2 + i\), \(2 - i\) from the structure, so the polynomial is \((x + 1)[x - (2 + i)][x - (2 - i)]\) (wait, maybe I misread earlier, let's correct: if roots are \(r_1, r_2, r_3\), then the polynomial is \((x - r_1)(x - r_2)(x - r_3)\). Wait, the first box was \((x + 1)(2 + ix)(2 - ix)\) – no, maybe the correct roots are -1, and then for the complex roots, if we have \(2 + i\) and \(2 - i\), then the factors should be \((x + 1)(x - (2 + i))(x - (2 - i))\). Wait, maybe the initial expression has a typo, but let's work with \((x + 1)(2 + ix)(2 - ix)\) – no, better to use the difference of squares for the complex part. Let's take the first expression: \((x + 1)(2 + ix)(2 - ix)\). First, multiply \((2 + ix)(2 - ix)\): that's \(2^2 - (ix)^2 = 4 - i^2x^2 = 4 + x^2\) (since \(i^2 = -1\)). Then multiply by \((x + 1)\): \((x + 1)(x^2 + 4) = x^3 + x^2 + 4x + 4\). Wait, but the polynomial is \(x^3 + Ax^2 + Bx + C\), so comparing, \(A = 1\), \(B = 4\), \(C = 4\)? Wait, no, maybe the roots are -1, and then the complex roots are from \((2 + ix)(2 - ix)\) – no, maybe the correct roots are -1, and then the quadratic factor is \(x^2 + 4\) (from \( (2 + ix)(2 - ix) = 4 + x^2 \)). So the polynomial is \((x + 1)(x^2 + 4) = x^3 + x^2 + 4x + 4\). So then \(A = 1\), \(B = 4\), \(C = 4\). Wait, let's check:

Step1: Multiply the complex factors

Take \((2 + ix)(2 - ix)\). Using \((a + b)(a - b) = a^2 - b^2\), here \(a = 2\), \(b = ix\). So:
\(2^2 - (ix)^2 = 4 - i^2x^2 = 4 + x^2\) (since \(i^2 = -1\))

Step2: Multiply by \((x + 1)\)

Now multiply \((x + 1)(x^2 + 4)\):
\(x(x^2 + 4) + 1(x^2 + 4) = x^3 + 4x + x^2 + 4 = x^3 + x^2 + 4x + 4\)

Step3: Compare with \(x^3 + Ax^2 + Bx + C\)

So \(A\) is the coefficient of \(x^2\), which is 1; \(B\) is the coefficient of \(x\), which is 4; \(C\) is the constant term, which is 4.

Wait, but maybe the roots are different. Let's re-examine the problem. The polynomial is of least degree with leading coefficient 1, so if the roots are -1, and the complex roots from \((2 + ix)(2 - ix)\) – no, maybe the roots are -1, and then the quadratic factor is \(x^2 + 4\), so the polynomial is \((x + 1)(x^2 + 4) = x^3 + x^2 + 4x + 4\). So:

Step1: Expand \((2 + ix)(2 - ix)\)

Using difference of squares: \( (2)^2 - (ix)^2 = 4 - i^2x^2 \). Since \(i^2 = -1\), this becomes \(4 + x^2\).

Step2: Multiply by \((x + 1)\)

\((x + 1)(x^2 + 4) = x(x^2 + 4) + 1(x^2 + 4) = x^3 + 4x + x^2 + 4 = x^3 + x^2 + 4x + 4\).

Step3: Identify coefficients

Compare with \(x^3 + Ax^2 + Bx + C\). So \(A = 1\), \(B = 4\), \(C = 4\).

Answer:

\(A = \boxed{1}\), \(B = \boxed{4}\), \(C = \boxed{4}\)