We need to simplify $(3x^2-2x+1)-(x^2+5x-4)$. The minus sign distributes to every term in the second polynomial: $3x^2-2x+1-x^2-5x+4$. Combining like terms: $x^2$ terms give $3x^2-x^2 = 2x^2$, $x$ terms give $-2x-5x = -7x$, and constants give $1+4 = 5$. The result is $2x^2-7x+5$. The most common error is forgetting to change the sign of the constant term -4 when distributing the minus sign, which would give an incorrect constant of -3. Remember: subtracting means adding the opposite of each term.

We need to find $k$ such that $2(x-3)+k = 2x+5$ for all $x$. First, distribute on the left side: $2(x-3) = 2x - 6$. So we have $2x - 6 + k = 2x + 5$. Since the $x$ terms are already equal on both sides, we need the constant terms to be equal: $-6 + k = 5$. Solving for $k$: $k = 5 + 6 = 11$. The key is recognizing that for two expressions to be equivalent for all values of $x$, both the coefficients of $x$ and the constant terms must match. Quick check: $2(x-3)+11 = 2x-6+11 = 2x+5$ ✓.

We need to simplify $-(x-2)^2+5x-1$. First, expand $(x-2)^2 = x^2-4x+4$. Applying the negative sign: $-(x^2-4x+4) = -x^2+4x-4$. Now we have $-x^2+4x-4+5x-1$. Combining like terms: the $x^2$ term is $-x^2$, the $x$ terms give $4x+5x = 9x$, and the constants give $-4-1 = -5$. The result is $-x^2+9x-5$. The key is to expand the squared term first, then apply the negative sign to all three terms. A common error is to forget the negative applies to all terms in the expansion.

We need to simplify $\frac{6x^2-9x}{3x}$. First, factor the numerator: $6x^2-9x = 3x(2x-3)$. This gives us $\frac{3x(2x-3)}{3x}$. Since $x \neq 0$, we can cancel the common factor $3x$ from numerator and denominator, leaving $2x-3$. The key is to factor completely before canceling. A common error is to cancel terms instead of factors, such as trying to cancel just the $x$ from $6x^2$ and $3x$. Always factor first, then cancel common factors.

We need to find $b$ such that $(x-5)^2 = x^2+bx+25$. Expanding the left side using the pattern $(a-b)^2 = a^2-2ab+b^2$: $(x-5)^2 = x^2 - 2(x)(5) + 25 = x^2 - 10x + 25$. Comparing with $x^2+bx+25$, we see that $b = -10$. The key is remembering that when squaring $(x-5)$, the middle term is negative: $-2 \cdot x \cdot 5 = -10x$. A common error is to think $b = 10$ by forgetting the negative sign in the expansion pattern.

We need to find $c$ such that $x^2+12x+c$ is a perfect square trinomial equal to $(x+6)^2$. Expanding $(x+6)^2$ using the pattern $(a+b)^2 = a^2+2ab+b^2$: $(x+6)^2 = x^2 + 2(x)(6) + 6^2 = x^2 + 12x + 36$. Comparing with $x^2+12x+c$, we see that $c = 36$. The key insight is that in a perfect square trinomial $x^2+bx+c$, if the middle coefficient is $b$, then $c = (b/2)^2$. Here, $b = 12$, so $c = (12/2)^2 = 6^2 = 36$.

To simplify $\frac{x^2-9}{x-3}$ for $x \neq 3$, first factor the numerator as a difference of squares: $x^2-9 = (x-3)(x+3)$. Now we have $\frac{(x-3)(x+3)}{x-3}$. Since $x \neq 3$, we can cancel the common factor $(x-3)$: $\frac{(x-3)(x+3)}{x-3} = x+3$. A common error is trying to cancel individual $x$ terms or the constant terms separately without first factoring, which is algebraically incorrect. Always factor completely before canceling common factors in rational expressions.

To factor $15x^2y-10xy^2$ completely, first find the GCF of both terms. The GCF of coefficients 15 and 10 is 5; the GCF of $x^2y$ and $xy^2$ is $xy$. So the GCF is $5xy$. Factoring out $5xy$: $15x^2y-10xy^2 = 5xy(3x-2y)$. We can verify: $5xy cdot 3x = 15x^2y$ and $5xy cdot (-2y) = -10xy^2$. A common error is factoring out only part of the GCF, like just $5x$ or $xy$, leaving common factors in the parentheses. Always factor out the complete GCF to get the simplest form.

To expand $(3t-2)(2t+5)$ using FOIL: First: $3t cdot 2t = 6t^2$; Outer: $3t cdot 5 = 15t$; Inner: $-2 cdot 2t = -4t$; Last: $-2 cdot 5 = -10$. Combining: $6t^2 + 15t - 4t - 10 = 6t^2 + 11t - 10$. The key is correctly handling the middle terms: $15t - 4t = 11t$, and getting the sign right on the constant: $-2 cdot 5 = -10$. A common error is adding the middle terms incorrectly or writing the last term as $+10$ instead of $-10$. Always double-check signs when multiplying negative numbers.

The expression $ax+6x-4$ needs to be rewritten as $(a+6)x+b$ to find $b$. First, combine the $x$ terms: $ax+6x = (a+6)x$. The constant term $-4$ remains unchanged. So we have $(a+6)x-4$, which means $b = -4$. A common error is thinking that $b$ somehow involves the variable $a$, or misreading the $-4$ as part of the $x$ term. When matching equivalent expressions, the constant term (the term without any variable) directly transfers to the new form.