We need to find $k$ such that $4(x-2)+k = 4x+5$ for all values of $x$. First, expand the left side: $4(x-2)+k = 4x-8+k$. For this to equal $4x+5$, we need $4x-8+k = 4x+5$. The $4x$ terms already match on both sides, so we need the constant terms to be equal: $-8+k = 5$. Solving for $k$: $k = 5+8 = 13$. A common error is forgetting to distribute the 4 to the -2, which would give $4x-2+k$ instead of $4x-8+k$. Always verify by substituting back: $4(x-2)+13 = 4x-8+13 = 4x+5$ ✓.

We need to factor $9a^2-25$, which is a difference of squares. The pattern for difference of squares is $A^2-B^2 = (A-B)(A+B)$. Here, $9a^2 = (3a)^2$ and $25 = 5^2$, so we have $(3a)^2-5^2$. Applying the formula: $(3a)^2-5^2 = (3a-5)(3a+5)$. The fully factored form is $(3a-5)(3a+5)$. A common mistake is not recognizing that $9a^2 = (3a)^2$ and trying to factor as $(9a-5)(a+5)$, which doesn't work. Always verify by expanding: $(3a-5)(3a+5) = 9a^2+15a-15a-25 = 9a^2-25$ ✓.

We need to simplify $2(x-3)^2-(x^2-6x+9)$. First, expand $(x-3)^2 = x^2-6x+9$. So we have $2(x^2-6x+9)-(x^2-6x+9)$. Distributing the 2 gives: $2x^2-12x+18-(x^2-6x+9)$. Now distribute the negative sign: $2x^2-12x+18-x^2+6x-9$. Combine like terms: $(2x^2-x^2)+(-12x+6x)+(18-9) = x^2-6x+9$. The key insight is recognizing that this equals $(x-3)^2$, and a common error is forgetting to distribute the negative sign to all terms in the second parentheses.

To expand $(2x-3)(x+5)$, we use FOIL method systematically. First terms: $2x cdot x = 2x^2$. Outer terms: $2x cdot 5 = 10x$. Inner terms: $-3 cdot x = -3x$. Last terms: $-3 cdot 5 = -15$. Combining all terms: $2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15$. The most common error is getting the sign wrong on the last term, writing $+15$ instead of $-15$ because students forget that $(-3)(+5) = -15$.

The question asks for the equivalent expression to (3(2x-5) + 4x) after combining like terms. Start by distributing the 3: (3 cdot 2x = 6x) and (3 cdot (-5) = -15), so the expression becomes (6x - 15 + 4x). Next, combine the like terms: (6x + 4x = 10x), and the constant is (-15), resulting in (10x - 15). A common error is forgetting to distribute the 3 fully, such as only multiplying the first term inside the parentheses, leading to incorrect combinations like (6x - 5 + 4x = 10x - 5). Another mistake might be changing the sign of the constant, resulting in (10x + 15). To verify, substitute a value like (x = 1) into the original and choices to check equivalence.

To factor $5a^2-20a$, we first identify the greatest common factor (GCF) of both terms. The GCF of $5a^2$ and $20a$ is $5a$ (since $5$ divides both coefficients and $a$ divides both variable parts). Factoring out $5a$ gives: $5a^2-20a = 5a(a) - 5a(4) = 5a(a-4)$. A common error is factoring out only part of the GCF, such as just $5$ or just $a$, which leaves a more complex expression in parentheses. Always factor out the complete GCF to get the simplest form.

We need to solve $2(3x+k)-5x = x+8$ for $k$. First, distribute on the left side: $6x + 2k - 5x = x + 8$. Combine like terms on the left: $(6x-5x) + 2k = x + 2k = x + 8$. Since the coefficients of $x$ on both sides are already equal (both are 1), we need the constant terms to match: $2k = 8$. Dividing both sides by 2 gives $k = 4$. The key insight is recognizing that once the $x$ terms match, the constant terms must also be equal.

To expand $(x-6)(x+2)$, we use the FOIL method or distributive property. First terms: $x cdot x = x^2$. Outer terms: $x cdot 2 = 2x$. Inner terms: $-6 cdot x = -6x$. Last terms: $-6 cdot 2 = -12$. Combining these gives $x^2 + 2x - 6x - 12 = x^2 - 4x - 12$. The most common error is incorrectly multiplying the signs, especially forgetting that $(-6)(2) = -12$, not $+12$.

To simplify $4(2t-1)-(3t-8)$, we first distribute the 4: $4(2t-1) = 8t-4$. Next, we need to subtract $(3t-8)$, which means changing the sign of each term inside: $-(3t-8) = -3t+8$. Now we have $8t-4-3t+8$. Combine like terms: $(8t-3t)+(-4+8) = 5t+4$. The most common error is forgetting that subtracting $(3t-8)$ changes both signs, incorrectly giving $-3t-8$ instead of $-3t+8$. Remember: subtracting a quantity means adding its opposite.

The task is to simplify $3(2x-5)-4(x+1)+2x$ by distributing and combining like terms. First, distribute the 3: $3(2x-5) = 6x - 15$. Next, distribute the -4: $-4(x+1) = -4x - 4$. Now combine all terms: $6x - 15 - 4x - 4 + 2x = (6x - 4x + 2x) + (-15 - 4) = 4x - 19$. A common error is distributing -4 as $-4x + 4$ instead of $-4x - 4$, which would incorrectly yield $4x - 11$. When distributing a negative number, both terms inside the parentheses change sign.