Question

which expression is equivalent to $x^2 + 2x + 2$? $(x + 1 - i)(x + 1 + i)$ $(x + 1 - i)(x + 1 - i)$ $(x + 2)(x + 1)$ $(x - 1 + i)(x - 1 - i)$

Explanation:

Step1: Recall the formula for factoring quadratic expressions with complex roots. The quadratic formula for roots of \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For \(x^2+2x + 2\), \(a = 1\), \(b = 2\), \(c = 2\). The discriminant is \(b^2-4ac=4 - 8=-4\). The roots are \(x=\frac{-2\pm\sqrt{-4}}{2}=\frac{-2\pm2i}{2}=-1\pm i\). So the quadratic can be factored as \((x - (-1 + i))(x - (-1 - i))=(x + 1 - i)(x + 1 + i)\).

Step2: Alternatively, expand each option:

• Option 1: \((x + 1 - i)(x + 1 + i)=[(x + 1)-i][(x + 1)+i]=(x + 1)^2 - i^2=x^2+2x + 1-(-1)=x^2+2x + 2\).
• Option 2: \((x + 1 - i)(x + 1 - i)=(x + 1 - i)^2=(x + 1)^2-2i(x + 1)+i^2=x^2+2x + 1-2ix-2i - 1=x^2+2x-2ix-2i

eq x^2+2x + 2\).

• Option 3: \((x + 2)(x + 1)=x^2+3x + 2

eq x^2+2x + 2\).

• Option 4: \((x - 1 + i)(x - 1 - i)=[(x - 1)+i][(x - 1)-i]=(x - 1)^2 - i^2=x^2-2x + 1-(-1)=x^2-2x + 2

eq x^2+2x + 2\).

Answer:

A. \((x + 1 - i)(x + 1 + i)\)