19. \( x^{6}-9 x^{4}-16 x^{2}+144 \)

To factor the polynomial \( x^{6} - 9x^{4} - 16x^{2} + 144 \), follow these steps: ### Step 1: Substitute \( y = x^2 \) This substitution simplifies the polynomial: \[ x^{6} - 9x^{4} - 16x^{2} + 144 \quad \rightarrow \quad y^3 - 9y^2 - 16y + 144 \] ### Step 2: Factor the Cubic Polynomial We need to factor \( y^3 - 9y^2 - 16y + 144 \). #### Find Rational Roots Possible rational roots are factors of 144: \[ \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm9, \pm12, \pm16, \pm18, \pm24, \pm36, \pm48, \pm72, \pm144 \] Testing \( y = 4 \): \[ 4^3 - 9 \cdot 4^2 - 16 \cdot 4 + 144 = 64 - 144 - 64 + 144 = 0 \] So, \( y = 4 \) is a root. Therefore, \( (y - 4) \) is a factor. #### Perform Polynomial Division Divide \( y^3 - 9y^2 - 16y + 144 \) by \( (y - 4) \): \[ y^3 - 9y^2 - 16y + 144 = (y - 4)(y^2 - 5y - 36) \] #### Factor the Quadratic Factor \( y^2 - 5y - 36 \): \[ y^2 - 5y - 36 = (y - 9)(y + 4) \] So, the factored form is: \[ y^3 - 9y^2 - 16y + 144 = (y - 4)(y - 9)(y + 4) \] ### Step 3: Substitute Back \( y = x^2 \) Replace \( y \) with \( x^2 \): \[ (x^2 - 4)(x^2 - 9)(x^2 + 4) \] ### Step 4: Factor Further if Possible Each quadratic can be factored: \[ x^2 - 4 = (x - 2)(x + 2) \\ x^2 - 9 = (x - 3)(x + 3) \\ x^2 + 4 \quad \text{(irreducible over the reals)} \] ### Final Factored Form Combining all factors: \[ (x - 2)(x + 2)(x - 3)(x + 3)(x^2 + 4) \] **Answer:** After factoring, the expression is (x – 2)(x + 2)(x – 3)(x + 3) multiplied by (x² + 4). Thus, (x−2)(x+2)(x−3)(x+3)(x²+4)