\[ P(x)=x^{4}+15 x^{2}-16 \] (a) Factor \( P \) into linear and irreducible quadratic factors with real coefficients. \( P(x)=\square \) (b) Factor \( P \) completely into linear factors with complex coefficients. \( P(x)= \)

To start with factoring the polynomial \( P(x) = x^{4} + 15x^{2} - 16 \) into linear and irreducible quadratic factors with real coefficients, we can make a substitution. Let \( y = x^{2} \), then the polynomial becomes: \[ P(y) = y^{2} + 15y - 16. \] Now we can factor this quadratic. We look for two numbers that multiply to \(-16\) (the constant term) and add to \(15\) (the coefficient of \(y\)). These numbers are \(16\) and \(-1\). Thus, we can factor \( P(y) \) as: \[ P(y) = (y + 16)(y - 1). \] Now, substituting back \( y = x^{2} \): \[ P(x) = (x^{2} + 16)(x^{2} - 1). \] The factor \( x^{2} - 1 \) can be further factored into: \[ x^{2} - 1 = (x + 1)(x - 1). \] Thus, the complete factorization into linear and irreducible quadratic factors with real coefficients is: \[ P(x) = (x^{2} + 16)(x + 1)(x - 1). \] For part (b), we need to factor \( P(x) \) completely into linear factors with complex coefficients. The factor \( x^{2} + 16 \) can be factored further using the fact that: \[ x^{2} + 16 = x^{2} + 4^2 = (x - 4i)(x + 4i). \] Putting this together, we get the complete factorization: \[ P(x) = (x - 4i)(x + 4i)(x + 1)(x - 1). \] So, the final factorizations are: (a) \( P(x) = (x^{2} + 16)(x + 1)(x - 1) \) (b) \( P(x) = (x - 4i)(x + 4i)(x + 1)(x - 1) \)