22. If $$ 24 x^{2}-k x+16=(3 x+4)(a x-b) $$ for all values of $$ x $$, where $$ a, b $$, and $$ k $$ are constants, what is the value of $$ k $$ ? $$ \quad 3 Q x^{2}-3 x b+4 a x-4 b $$

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The value of $$ k $$ is -44. To solve for $$ k $$ in the equation $$ 24 x^{2}-k x+16=(3 x+4)(a x-b) $$, we first expand the right-hand side. Starting with the expression: $$ (3x + 4)(ax - b) $$ We can use the distributive property (also known as the FOIL method for binomials): $$ = 3x(ax) + 3x(-b) + 4(ax) + 4(-b) $$ $$ = 3ax^2 - 3bx + 4ax - 4b $$ Now, we can combine the like terms: $$ = 3ax^2 + (4a - 3b)x - 4b $$ Next, we set this equal to the left-hand side of the original equation: $$ 24x^2 - kx + 16 = 3ax^2 + (4a - 3b)x - 4b $$ Now, we can equate the coefficients of the corresponding powers of $$ x $$: 1. For $$ x^2 $$: $$ 3a = 24 $$ Solving for $$ a $$: $$ a = \frac{24}{3} = 8 $$ 2. For $$ x $$: $$ 4a - 3b = -k $$ Substituting $$ a = 8 $$: $$ 4(8) - 3b = -k $$ $$ 32 - 3b = -k $$ 3. For the constant term: $$ -4b = 16 $$ Solving for $$ b $$: $$ b = -4 $$ Now, substituting $$ b = -4 $$ back into the equation for $$ k $$: $$ 32 - 3(-4) = -k $$ $$ 32 + 12 = -k $$ $$ 44 = -k $$ $$ k = -44 $$ Thus, the value of $$ k $$ is $$ \boxed{-44} $$

22. If \( 24 x^{2}-k x+16=(3 x+4)(a x-b) \) for all values of \( x \), where \( a, b \), and \( k \) are constants, what is the value of \( k \) ? \( \quad 3 Q x^{2}-3 x b+4 a x-4 b \)

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22. If \( 24 x^{2}-k x+16=(3 x+4)(a x-b) \) for all values of \( x \), where \( a, b \), and \( k \) are constants, what is the value of \( k \) ? \( \quad 3 Q x^{2}-3 x b+4 a x-4 b \)

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