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Factor completely:

(2x+3)(4x+9)+(5x+4)(4x+9)
Answer:

Factor completely: $\newline$ $(2 x+3)(4 x+9)+(5 x+4)(4 x+9)$ $\newline$ Answer:

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Powers with negative bases

Full solution

Q. Factor completely:

\newline

(2 x+3)(4 x+9)+(5 x+4)(4 x+9)

\newline

Answer:

Identify Common Factors: Identify common factors in both terms. $\newline$ We have two terms: \(2x+ $3$ )( $4$ x+ $9$ )\ and \(5x+ $4$ )( $4$ x+ $9$ )\. Both terms have a common factor of \(4x+ $9$ )\.
Factor Out Common Factor: Factor out the common factor $(4x+9)$ . We can write the expression as $(4x+9)$ being multiplied by the sum of the other factors from each term, which are $(2x+3)$ and $(5x+4)$ . So, $(2x+3)(4x+9)+(5x+4)(4x+9) = (4x+9)((2x+3)+(5x+4))$ .
Simplify Inside Parentheses: Simplify the expression inside the parentheses. $\newline$ Now we need to add the terms $(2x+3)$ and $(5x+4)$ together. $\newline$ $(2x+3) + (5x+4) = 2x + 3 + 5x + 4 = (2x + 5x) + (3 + 4) = 7x + 7$ .
Write Final Factored Form: Write the final factored form. $\newline$ Now we can write the completely factored form of the original expression by combining the common factor $(4x+9)$ with the simplified sum $(7x+7)$ . $\newline$ So, $(4x+9)((2x+3)+(5x+4)) = (4x+9)(7x+7)$ .

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