Question

multiply vertically.$(x^{2}+3x+9)(9x^{2}-9x+5)$$(x^{2}+3x+9)(9x^{2}-9x+5) = \square$ (simplify your answer.)

Explanation:

Step1: Multiply by 5

$(x^2 + 3x + 9) \times 5 = 5x^2 + 15x + 45$

Step2: Multiply by $-9x$

$(x^2 + 3x + 9) \times (-9x) = -9x^3 - 27x^2 - 81x$

Step3: Multiply by $9x^2$

$(x^2 + 3x + 9) \times 9x^2 = 9x^4 + 27x^3 + 81x^2$

Step4: Sum all products

$9x^4 + 27x^3 + 81x^2 - 9x^3 - 27x^2 - 81x + 5x^2 + 15x + 45$

Step5: Combine like terms

$9x^4 + (27x^3 - 9x^3) + (81x^2 - 27x^2 + 5x^2) + (-81x + 15x) + 45$
$=9x^4 + 18x^3 + 59x^2 - 66x + 45$

Answer:

$9x^4 + 18x^3 + 59x^2 - 66x + 45$