Algebraic Techniques

$(2x - 3)^2 = (2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9$. The student squared the two terms but dropped the middle term $-12x$ (and the sign of the constant). Remember $(a - b)^2 \ne a^2 - b^2$.

FOIL: $2x \times x = 2x^2$, $2x \times 4 = 8x$, $-3 \times x = -3x$, $-3 \times 4 = -12$. Combine: $2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$.

$(x+4)(x-4) = x^2 - 16$ and $(x-2)^2 = x^2 - 4x + 4$. Subtract, distributing the minus: $x^2 - 16 - (x^2 - 4x + 4) = x^2 - 16 - x^2 + 4x - 4 = 4x - 20$.

First take out the common factor 3: $3(x^2 - 4)$. Then $x^2 - 4$ is a difference of two squares: $(x - 2)(x + 2)$. So the full factorisation is $3(x - 2)(x + 2)$. ($3(x^2 - 4)$ is not fully factorised.)

Since $(a + b)^2 = a^2 + 2ab + b^2$, the statement holds only when the middle term $2ab = 0$, i.e. when $a = 0$ or $b = 0$. For all other values it is false.

Area is length $\times$ width: $(x + 7)(x + 2) = x^2 + 2x + 7x + 14 = x^2 + 9x + 14$. ($2x + 18$ is the perimeter, not the area.)

FOIL: $x^2$, $x \times -3 = -3x$, $5 \times x = 5x$, $5 \times -3 = -15$: $x^2 - 3x + 5x - 15 = x^2 + 2x - 15$. ($x^2 - 2x - 15$ got the middle sign wrong; $x^2 + 8x - 15$ added 5 and 3 ignoring signs.)

Length $= \dfrac{\text{area}}{\text{width}}$. Factorise the area: $x^2 + 7x + 12 = (x + 3)(x + 4)$, so $\dfrac{(x + 3)(x + 4)}{x + 3} = x + 4$ cm.

Perfect square $(a + b)^2 = a^2 + 2ab + b^2$ with $a = x$ and $b = 5$: $x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25$. ($x^2 + 25$ is the canonical error: it omits the middle term $2ab = 10x$. Remember $(a + b)^2 \ne a^2 + b^2$.)

Write the product as $(50 + 3)(50 - 3) = 50^2 - 3^2 = 2500 - 9 = 2491$. This is the identity $(a + b)(a - b) = a^2 - b^2$ used on numbers instead of letters.

The HCF of 6 and 9 is 3: $6x + 9 = 3 \times 2x + 3 \times 3 = 3(2x + 3)$. ($3(2x + 9)$ forgot to divide the 9; $6(x + 1.5)$ is not valid because $1.5$ is not a whole number, so 6 is not the HCF.)

The perimeter is $2(\text{length} + \text{width})$, so $\text{length} + \text{width} = \dfrac{10x + 6}{2} = 5x + 3$. Then the width $= (5x + 3) - (3x + 4) = 2x - 1$ cm.

The numerical HCF of 12 and 8 is 4 and the variable HCF of $x^2$ and $x$ is $x$, so the HCF is $4x$: $12x^2 - 8x = 4x \times 3x - 4x \times 2 = 4x(3x - 2)$. ($4(3x^2 - 2x)$ took only the numerical HCF; $4x(3x - 8)$ did not divide $-8x$ by $4x$.)

$x^2 - 2x - 15 = (x - 5)(x + 3)$. Substituting $x = 5$: $(5 - 5)(5 + 3) = 0 \times 8 = 0$, so $x = 5$ is a root of the expression.

The numerical HCF of 8 and 12 is 4 and the variable HCF is $xy$, so the HCF is $4xy$: $8x^2y - 12xy^2 = 4xy \times 2x - 4xy \times 3y = 4xy(2x - 3y)$. ($4xy(2x - 12y)$ did not divide $-12xy^2$ correctly; $4(2x^2y - 3xy^2)$ missed the $xy$.)

Find two numbers that multiply to $+6$ and add to $+5$: that is 2 and 3, both positive, so $(x + 2)(x + 3)$. ($(x + 6)(x + 1)$ multiplies to 6 but adds to 7; $(x + 5)(x + 6)$ just copied the coefficients.)

Find two numbers that multiply to $-6$ and add to $+1$: that is $+3$ and $-2$, so $(x + 3)(x - 2)$. ($(x - 3)(x + 2)$ reverses the signs, giving $x^2 - x - 6$; $(x + 6)(x - 1)$ sums to $+5$.)

The curve meets the $x$-axis where $y = 0$. Factorise: $x^2 - 5x + 6 = (x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$.

The product is $-12$ and the sum is $-1$, so the larger value is negative: $-4$ and $+3$. So $(x - 4)(x + 3)$. ($(x + 4)(x - 3)$ reverses the signs, giving $x^2 + x - 12$; $(x - 12)(x + 1)$ sums to $-11$.)

This is a difference of two squares: $x^2 - 25 = x^2 - 5^2 = (x + 5)(x - 5)$. ($(x - 5)^2 = x^2 - 10x + 25$ has a middle term, so it is not $x^2 - 25$; the expression certainly is factorisable.)

Simplify the numbers and the variables separately: $\dfrac{3}{6} = \dfrac{1}{2}$ and $\dfrac{x^2}{x} = x$, so $\dfrac{3x^2}{6x} = \dfrac{x}{2}$. ($\dfrac{3x}{6}$ did not cancel the variable; $\dfrac{1}{2x}$ cancelled the variable wrongly.)

Same denominator, so subtract the numerators, distributing the minus across the whole second numerator: $(x + 2) - (x - 1) = x + 2 - x + 1 = 3$, so the result is $\dfrac{3}{3} = 1$. ($\dfrac{1}{3}$ is the canonical error: it failed to distribute the minus, writing $x + 2 - x - 1$ instead of $x + 2 - x + 1$.)

The mean is the sum divided by 2: $\dfrac{(x + 5) + (x - 1)}{2} = \dfrac{2x + 4}{2} = x + 2$. ($2x + 4$ forgot to divide by 2.)

First expand $(x + 3)^2 = x^2 + 6x + 9$, then add $2x$: $x^2 + 6x + 9 + 2x = x^2 + 8x + 9$ (since $6x + 2x = 8x$). ($x^2 + 6x + 9$ forgot to add the $2x$; $x^2 + 9 + 2x$ did not expand the middle term of the perfect square.)

Factorise the numerator as a difference of two squares: $x^2 - 9 = (x + 3)(x - 3)$. Cancel the common factor $(x - 3)$: $\dfrac{(x + 3)(x - 3)}{x - 3} = x + 3$ (for $x \ne 3$). ($x - 3$ cancelled the wrong factor; $x^2 - 6$ subtracted instead of dividing.)