Factor completely: 9u^(2)+21 us^(3)+10s^(6) Answer:

Identify Common Factors: Look for common factors in each term of the polynomial 9u^2 + 21us^3 + 10s^6. We notice that there are no common factors among all three terms.Search for Patterns: Since there are no common factors, we look for patterns or special products such as a difference of squares, perfect square trinomials, or sum/difference of cubes. However, none of these patterns are present in this polynomial.Attempt Factor by Grouping: We attempt to factor by grouping. To do this, we need to rearrange the terms or find a grouping that works, but with the given terms 9u^2, 21us^3, and 10s^6, there is no clear way to group them to factor by grouping.Conclude in Simplest Form: Since the polynomial does not have a common factor, does not fit a special product pattern, and cannot be factored by grouping, we conclude that the polynomial is already in its simplest form and cannot be factored further over the integers.

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Factor completely:

9u^(2)+21 us^(3)+10s^(6)
Answer:

Factor completely: $\newline$ $9 u^{2}+21 u s^{3}+10 s^{6}$ $\newline$ Answer:

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Algebra 2

Composition of linear and quadratic functions: find a value

Full solution

Q. Factor completely:

\newline

9 u^{2}+21 u s^{3}+10 s^{6}

\newline

Answer:

Identify Common Factors: Look for common factors in each term of the polynomial $9u^2 + 21us^3 + 10s^6$ . We notice that there are no common factors among all three terms.
Search for Patterns: Since there are no common factors, we look for patterns or special products such as a difference of squares, perfect square trinomials, or sum/difference of cubes. However, none of these patterns are present in this polynomial.
Attempt Factor by Grouping: We attempt to factor by grouping. To do this, we need to rearrange the terms or find a grouping that works, but with the given terms $9u^2$ , $21us^3$ , and $10s^6$ , there is no clear way to group them to factor by grouping.
Conclude in Simplest Form: Since the polynomial does not have a common factor, does not fit a special product pattern, and cannot be factored by grouping, we conclude that the polynomial is already in its simplest form and cannot be factored further over the integers.