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.

We need to find $k$ such that $2(x-3)+5x=7x+k$ for all values of $x$. First, distribute on the left side: $2(x-3) = 2x - 6$. The left side becomes $2x - 6 + 5x = 7x - 6$. For the equation $7x - 6 = 7x + k$ to be true for all $x$, the coefficients of $x$ must match (they do: both are 7) and the constant terms must match. Therefore, $-6 = k$, so $k = -6$. A common mistake is to write $k = 6$ by forgetting the negative sign.

We need to expand $(x-6)(x+2)$ and identify the coefficient $b$ in the standard form $x^2+bx+c$. Using FOIL: First terms give $x cdot x = x^2$, Outer terms give $x cdot 2 = 2x$, Inner terms give $-6 cdot x = -6x$, and Last terms give $-6 cdot 2 = -12$. Combining these: $x^2 + 2x - 6x - 12 = x^2 - 4x - 12$. The coefficient of $x$ is $b = -4$. A common mistake is adding the inner and outer products incorrectly, getting $2x + 6x = 8x$ instead of $2x - 6x = -4x$.

We need to find $m$ such that $x^2+12x+36 = (x+m)^2$. Expanding the right side: $(x+m)^2 = x^2+2mx+m^2$. Comparing coefficients with $x^2+12x+36$, we need $2m = 12$ (coefficient of $x$) and $m^2 = 36$ (constant term). From $2m = 12$, we get $m = 6$. Let's verify: if $m = 6$, then $m^2 = 36$ ✓, confirming our answer. Therefore, $x^2+12x+36 = (x+6)^2$ and $m = 6$. This is a perfect square trinomial, and recognizing the pattern $a^2+2ab+b^2 = (a+b)^2$ helps identify $m$ quickly.

We need to simplify $(2x+3)(x-4) - (x-4)$. Notice that $(x-4)$ is a common factor. We can rewrite this as $(2x+3)(x-4) - 1(x-4) = (x-4)[(2x+3)-1] = (x-4)(2x+3-1) = (x-4)(2x+2)$. The expression simplifies to $(x-4)(2x+2)$. Alternatively, we could expand everything first: $(2x+3)(x-4) = 2x^2-8x+3x-12 = 2x^2-5x-12$, then subtract $(x-4)$ to get $2x^2-5x-12-x+4 = 2x^2-6x-8$, which factors as $2(x^2-3x-4) = 2(x-4)(x+1) = (x-4)(2x+2)$. The factored form $(x-4)(2x+2)$ is simpler than the expanded form.

We need to factor $5x^2-20x$ by finding the greatest common factor (GCF). Looking at both terms, $5x^2 = 5 cdot x cdot x$ and $20x = 5 cdot 4 cdot x$, so the GCF is $5x$. Factoring out $5x$: $5x^2-20x = 5x(x) - 5x(4) = 5x(x-4)$. The expression in factored form is $5x(x-4)$. A common error is factoring out only $5$ or only $x$ instead of the full GCF $5x$, which would give incomplete factorization. Always check your factoring by distributing back to verify you get the original expression.

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 $-4(x+1) = -4x-4$. Now we have $6x-15-4x-4+7$. Combine the $x$ terms: $6x-4x = 2x$. Combine the constants: $-15-4+7 = -12$. Therefore, the expression simplifies to $2x-12$. A common error is forgetting to distribute the negative sign with $-4$, which would incorrectly give $-4x+4$ instead of $-4x-4$.

To factor $4x^2 - 9$, recognize this as a difference of squares since $4x^2 = (2x)^2$ and $9 = 3^2$. Using the pattern $a^2 - b^2 = (a-b)(a+b)$, we get $4x^2 - 9 = (2x)^2 - 3^2 = (2x-3)(2x+3)$. To verify, expand: $(2x-3)(2x+3) = 4x^2 + 6x - 6x - 9 = 4x^2 - 9$ ✓. A common mistake is factoring as $(4x-3)(x+3)$, but this expands to $4x^2 + 12x - 3x - 9 = 4x^2 + 9x - 9$, which has an unwanted middle term.