Find each product. 17) (4m-3)(8m+1) 19) (5n-4)(4n^2+4n-1) 18) (7x+5)(4x-4) 20) (n-6)(n^2+7n+4)

17) To find the product of $(4m-3)(8m+1)$, we can use the distributive property (also known as the FOIL method for binomials). <br /><br />First, multiply the first terms: $4m \cdot 8m = 32m^2$.<br /><br />Next, multiply the outer terms: $4m \cdot 1 = 4m$.<br /><br />Then, multiply the inner terms-3 \cdot 8m = -24m$.<br /><br />Finally, multiply the last terms: $-3 \cdot 1 = -3$.<br /><br />Now, combine like terms: $32m^2 + 4m - 24m - 3 = 32m^2 - 20m - 3$.<br /><br />So, the product of $(4m-3)(8m+1)$ is $32m^2 - 20m - 3$.<br /><br />19) To find the product of $(5n-4)(4n^2+4n-1)$, we can use the distributive property.<br /><br />First, multiply the first term of the first binomial by each term in the second trinomial: $5n \cdot 4n^2 = 20n^3$, $5n \cdot 4n = 20n^2$, $5n \cdot (-1) = -5n$.<br /><br />Next, multiply the second term of the first binomial by each term in the second trinomial: $-4 \cdot 4n^2 = -16n^2$, $-4 \cdot 4n = -16n$, $-4 \cdot (-1) = 4$.<br /><br />Now, combine like terms: $20n^3 + 20n^2 - 5n - 16n^2 - 16n + 4 = 20n^3 + 4n^2 - 21n + 4$.<br /><br />So, the product of $(5n-4)(4n^2+4n-1)$ is $20n^3 + 4n^2 - 21n + 4$.<br /><br />18) To find the product of $(7x+5)(4x-4)$, we can use the distributive property.<br /><br />First, multiply the first terms: $7x \cdot 4x = 28x^2$.<br /><br />Next, multiply the outer terms: $7x \cdot (-4) = -28x$.<br /><br />Then, multiply the inner terms: $5 \cdot 4x = 20x$.<br /><br />Finally, multiply the last terms: $5 \cdot (-4) = -20$.<br /><br />Now, combine like terms: $28x^2 - 28x + 20x - 20 = 28x^2 - 8x - 20$.<br /><br />So, the product of $(7x+5)(4x-4)$ is $28x^2 - 8x - 20$.<br /><br />20) To find the product of $(n-6)(n^2+7n+4)$, we can use the distributive property.<br /><br />First, multiply the first term of the first binomial by each term in the second trinomial: $n \cdot n^2 = n^3$, $n \cdot 7n = 7n^2$, $n \cdot 4 = 4n$.<br /><br />Next, multiply the second term of the first binomial by each term in the second trinomial: $-6 \cdot n^2 = -6n^2$, $-6 \cdot 7n = -42n$, $-6 \cdot 4 = -24$.<br /><br />Now, combine like terms: $n^3 + 7n^2 + 4n - 6n^2 - 42n - 24 = n^3 + n^2 - 38n - 24$.<br /><br />So, the product of $(n-6)(n^2+7n+4)$ is $n^3 + n^2 - 38n - 24$.