We need to find $a$ such that $x^2+14x+49 = (x+a)^2$. Expanding $(x+a)^2 = x^2+2ax+a^2$, we need $2a = 14$ and $a^2 = 49$. From $2a = 14$, we get $a = 7$, and checking: $7^2 = 49$ ✓. This is a perfect square trinomial where the middle coefficient is twice the product of the square roots. Common errors include using $a = 14$ (not halving the middle coefficient) or $a = 49$ (using the constant term directly). Remember that in $(x+a)^2$, the middle term is $2ax$, not $ax$.

We need to expand $(3x+2)(x-5)$ using FOIL (First, Outer, Inner, Last). First: $3x cdot x = 3x^2$. Outer: $3x cdot (-5) = -15x$. Inner: $2 cdot x = 2x$. Last: $2 cdot (-5) = -10$. Combining: $3x^2-15x+2x-10 = 3x^2-13x-10$. Common mistakes include dropping the negative sign on $-15x$ (giving $3x^2+15x+2x-10$) or miscalculating the constant term as positive 10. Always check signs carefully when multiplying.

We need to find $k$ such that $2(3x+k)-5x$ equals $x+8$ for all values of $x$. First, expand the left side: $2(3x+k)-5x = 6x+2k-5x = x+2k$. For this to equal $x+8$, we need the coefficients of $x$ and the constants to match separately. The $x$ coefficients already match (both are 1), and the constants give us $2k = 8$, so $k = 4$. A common error is confusing which terms to match or solving $6x+2k-5x = 8$ incorrectly by mixing the $x$ terms with constants.

We need to simplify $5(2y-3)-(y-4)+2y$ by carefully distributing and combining like terms. First, distribute: $5(2y-3) = 10y-15$. For $-(y-4)$, the negative sign changes both terms inside: $-(y-4) = -y+4$. Now we have $10y-15-y+4+2y$. Combining like terms: $y$ terms give $10y-y+2y = 11y$, and constants give $-15+4 = -11$. Therefore, the expression equals $11y-11$. The key error to avoid is writing $-(y-4)$ as $-y-4$ instead of $-y+4$.

We need to factor $4x^2-25$, which is a difference of squares since $4x^2 = (2x)^2$ and $25 = 5^2$. The difference of squares pattern is $a^2-b^2 = (a-b)(a+b)$, so $4x^2-25 = (2x)^2-5^2 = (2x-5)(2x+5)$. A common mistake is factoring out only a numerical GCF or trying to write it as $(4x-5)(x+5)$, which when expanded gives $4x^2+20x-5x-25 = 4x^2+15x-25$, not our original expression. Always check your factorization by expanding: $(2x-5)(2x+5) = 4x^2+10x-10x-25 = 4x^2-25$ ✓.

We need to simplify $3(2x-5)-4(x+1)+7$ by distributing and combining like terms. First, distribute: $3(2x-5) = 6x-15$ and $-4(x+1) = -4x-4$ (note the negative sign applies to both terms). This gives us $6x-15-4x-4+7$. Combining like terms: $x$ terms give $6x-4x = 2x$, and constants give $-15-4+7 = -12$. Therefore, the expression simplifies to $2x-12$. A common error is forgetting to distribute the negative sign, which would incorrectly yield $-4x+4$ instead of $-4x-4$.

We need to simplify $\frac{2x^2-8}{2x}$ for $x \neq 0$. First, factor the numerator: $2x^2-8 = 2(x^2-4)$. Now we have $\frac{2(x^2-4)}{2x} = \frac{x^2-4}{x}$. We can separate this fraction: $\frac{x^2-4}{x} = \frac{x^2}{x} - \frac{4}{x} = x - \frac{4}{x}$. A common error is trying to cancel terms across addition/subtraction before factoring, or factoring $x^2-4$ further as $(x-2)(x+2)$ which isn't necessary here. The key is recognizing that we can split the fraction after simplifying the common factor of 2.

This question asks to simplify (5y - 2(3y - 4) + 7) by distributing and combining like terms. Begin by distributing the -2: (-2 cdot 3y = -6y) and (-2 cdot (-4) = 8), so the expression becomes (5y - 6y + 8 + 7). Now combine like terms: (5y - 6y = -y) for the y-terms, and (8 + 7 = 15) for the constants, resulting in (-y + 15). A common error is forgetting the positive 8 from distributing the negative to -4, leading to incorrect constants like +7 or -1. Another mistake involves not distributing the negative sign properly, such as treating it as subtraction without flipping the signs inside. When dealing with negative distributions, double-check the signs by expanding step-by-step to ensure accuracy.

The question asks for the fully simplified equivalent expression to (3(2x-5) + 4(x+1)) by distributing and combining like terms. Start by distributing the 3: (3 cdot 2x = 6x) and (3 cdot (-5) = -15), resulting in (6x - 15). Next, distribute the 4: (4 cdot x = 4x) and (4 cdot 1 = 4), so add (4x + 4) to get (6x - 15 + 4x + 4). Combine the like terms: (6x + 4x = 10x) and (-15 + 4 = -11), yielding (10x - 11). A common error is mishandling the constants, such as adding -15 and 4 as -19 instead of -11. Another mistake might be forgetting to distribute fully, leaving it as (6x - 15 + 4x + 1), which is not simplified. When simplifying expressions, always distribute first and then group like terms to avoid sign errors.

The question requires factoring $(x^2$ - 9) to find an equivalent expression representing possible side lengths of a rectangle with that area. Recognize that $(x^2$ - 9) is a difference of squares, which factors as ((x + a)(x - a)) where $(a^2$ = 9), so (a = 3). Thus, it factors to ((x + 3)(x - 3)). Verify by expanding: (x cdot x = $x^2$), (x cdot (-3) = -3x), (3 cdot x = 3x), and (3 cdot (-3) = -9), combining to $(x^2$ + (-3x + 3x) - 9 = $x^2$ - 9). A key error is confusing it with a perfect square like ((x - $3)^2$ = $x^2$ - 6x + 9), which adds a middle term. Another mistake is uneven factoring, such as (x(x - 9) = $x^2$ - 9x), missing the constant. For factoring questions, check your answer by expanding back to the original to confirm equivalence.