To find the product of \((2q^9 + 3q^7)(-6q^2 + 9)\), we can use the distributive property (also known as the FOIL method for binomials). We will multiply each term in the first polynomial by each term in the second polynomial.
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Multiply \(2q^9\) by \(-6q^2\): \[ 2q^9 \cdot -6q^2 = -12q^{9+2} = -12q^{11} \]
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Multiply \(2q^9\) by \(9\): \[ 2q^9 \cdot 9 = 18q^9 \]
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Multiply \(3q^7\) by \(-6q^2\): \[ 3q^7 \cdot -6q^2 = -18q^{7+2} = -18q^{9} \]
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Multiply \(3q^7\) by \(9\): \[ 3q^7 \cdot 9 = 27q^7 \]
Now we sum all these products together: \[ -12q^{11} + 18q^9 - 18q^9 + 27q^7 \]
Combine like terms:
- The \(q^9\) terms: \[ 18q^9 - 18q^9 = 0 \]
- The final expression becomes: \[ -12q^{11} + 27q^7 \]
Thus, the product of \((2q^9 + 3q^7)(-6q^2 + 9)\) is: \[ -12q^{11} + 27q^{7} \]
So the answer is: \[ \boxed{-12q^{11} + 27q^7} \]