Question

1. factor the binomial.

x³+81

○ (x+9)(x²−3x+27)

○ the polynomial cannot be factored.

○ (x−3)(x²+9x+27)

○ (x+3)(x²−9x+27)

Explanation:

To factor \(x^{3}+81\), we check if it fits sum of cubes formula \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\). Rewrite \(81\) as \(3^{3}\times3=27\times3\), not a perfect cube. So \(x^{3}+81=x^{3}+3^{3}\times3\), which doesn't match sum/difference of cubes (needs \(b^{3}\) where \(b\) is integer cube root). Testing given options:

• For \((x + 9)(x^{2}-3x + 27)\): Expand \((x + 9)(x^{2}-3x + 27)=x^{3}-3x^{2}+27x+9x^{2}-27x + 243=x^{3}+6x^{2}+243

eq x^{3}+81\).

• For \((x - 3)(x^{2}+9x + 27)\): Expand \((x - 3)(x^{2}+9x + 27)=x^{3}+9x^{2}+27x-3x^{2}-27x - 81=x^{3}+6x^{2}-81

eq x^{3}+81\).

• For \((x + 3)(x^{2}-9x + 27)\): Expand \((x + 3)(x^{2}-9x + 27)=x^{3}-9x^{2}+27x+3x^{2}-27x + 81=x^{3}-6x^{2}+81

eq x^{3}+81\).

Thus, \(x^{3}+81\) cannot be factored with integer coefficients as given options fail, and it doesn't fit standard factoring forms.

Answer:

The polynomial cannot be factored.