Write the polynomial as a product of linear factors. \[ x^{4}-5 x^{2}-36 \] \( (x-[?])(x+3)(x-[\square i)(x+[] i) \)

To factor the polynomial \( x^{4}-5 x^{2}-36 \) into linear factors, we can start by making a substitution to simplify the expression. Let's set \( y = x^2 \). Then, the polynomial becomes: \[ y^2 - 5y - 36 \] Next, we will factor this quadratic expression. We need to find two numbers that multiply to \(-36\) (the constant term) and add to \(-5\) (the coefficient of \(y\)). The numbers that satisfy these conditions are \( -9 \) and \( 4 \) because: \[ -9 \times 4 = -36 \] \[ -9 + 4 = -5 \] Thus, we can factor the quadratic as: \[ (y - 9)(y + 4) \] Now, substituting back \( y = x^2 \), we have: \[ (x^2 - 9)(x^2 + 4) \] Next, we can factor \( x^2 - 9 \) further, as it is a difference of squares: \[ (x - 3)(x + 3) \] The term \( x^2 + 4 \) cannot be factored into real linear factors, but it can be factored into complex linear factors: \[ x^2 + 4 = (x - 2i)(x + 2i) \] Putting it all together, we have: \[ x^{4}-5 x^{2}-36 = (x - 3)(x + 3)(x - 2i)(x + 2i) \] Thus, the polynomial can be expressed as a product of linear factors: \[ (x - 3)(x + 3)(x - 2i)(x + 2i) \] So, filling in the blanks, we have: \[ (x - [3])(x + 3)(x - [2i])(x + [2i]) \]