Algebraic Techniques (Trinomials & Identities)

$(2x - 3)^2 = (2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9$. The student kept only the squared terms (as if it were a difference of squares) and dropped the middle term $-12x$; the correct answer is $4x^2 - 12x + 9$.

Use the common denominator $x(x + 1)$: $\dfrac{2(x + 1) + 3x}{x(x + 1)} = \dfrac{5x + 2}{x(x + 1)}$. (You cannot just add numerators and denominators separately.)

Take out the common factor $3x$ first: $3x(x^2 - 4)$, then difference of squares: $3x(x - 2)(x + 2)$. ($3x(x^2 - 4)$ is only partly factorised.)

Group in pairs: $2a(x - 3) + b(x - 3)$, then take out the common factor $(x - 3)$: $(x - 3)(2a + b)$.

FOIL: $(2x)(x) + (2x)(4) + (3)(x) + (3)(4) = 2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12$. ($2x^2 + 12x + 11$ swapped the middle coefficient and constant; $2x^2 + 12$ forgot the middle terms.)

$(3x + 2)^2 = (3x)^2 + 2(3x)(2) + 2^2 = 9x^2 + 12x + 4$. Square the whole $3x$, giving $9x^2$. ($3x^2 + 12x + 4$ did not square the 3; $9x^2 + 4$ forgot the middle term.)

$x(x^2 + 2x + 4) - 2(x^2 + 2x + 4) = x^3 + 2x^2 + 4x - 2x^2 - 4x - 8 = x^3 - 8$ (the middle terms cancel; this is the difference of cubes).

Write the product as a difference of squares: $103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2 = 10000 - 9 = 9991$. (This is the identity $(a + b)(a - b) = a^2 - b^2$ used on numbers instead of letters.)

Width $= \dfrac{\text{area}}{\text{length}}$. Factorise the area: $x^2 + 7x + 10 = (x + 5)(x + 2)$, so $\dfrac{(x + 5)(x + 2)}{x + 5} = x + 2$ cm.

The remaining area is a difference of two squares: $(2x + 1)^2 - (x + 2)^2$. Factorise directly: $[(2x + 1) + (x + 2)][(2x + 1) - (x + 2)] = (3x + 3)(x - 1) = 3(x + 1)(x - 1)$.

Two numbers that multiply to 12 and add to 7 are 3 and 4, so $(x + 3)(x + 4)$. ($(x + 2)(x + 6)$ adds to 8; $(x - 3)(x - 4)$ gives $x^2 - 7x + 12$.)

The curve meets the $x$-axis where $y = 0$. Factorise: $x^2 - 7x + 12 = (x - 3)(x - 4) = 0$, so $x = 3$ or $x = 4$.

Square both sides: $\left(x + \dfrac{1}{x}\right)^2 = x^2 + 2 + \dfrac{1}{x^2} = 9$, so $x^2 + \dfrac{1}{x^2} = 9 - 2 = 7$.

Non-monic: split the middle using $2 \times 3 = 6$ with factors $6$ and $1$ (sum $7$): $2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$. ($(2x + 3)(x + 1)$ expands to $2x^2 + 5x + 3$.)

Difference of squares twice: $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$. ($x^2 + 4$ does not factor further over the reals; $(x^2 - 4)(x^2 + 4)$ is only partly done.)

$(a + b)^2 = a^2 + 2ab + b^2$, which equals $a^2 + b^2$ exactly when the middle term $2ab = 0$, i.e. when $a = 0$ or $b = 0$.

Volume $= \text{base area} \times \text{height} = (x + 1)^2 \times x = x(x^2 + 2x + 1) = x^3 + 2x^2 + x$.

Non-monic: $6 \times (-2) = -12$ with factors $-4$ and $3$ (sum $-1$): $6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (3x - 2)(2x + 1)$. ($(3x + 2)(2x - 1)$ expands to $6x^2 + x - 2$.)

Square the sum: $(a + b)^2 = a^2 + 2ab + b^2 = 49$. Subtract $2ab$: $a^2 + b^2 = 49 - 2(12) = 49 - 24 = 25$.

By Pythagoras, $x^2 + (x + 7)^2 = (x + 8)^2$. Expanding gives $x^2 + x^2 + 14x + 49 = x^2 + 16x + 64$, which simplifies to $x^2 - 2x - 15 = 0$. Factorise: $(x - 5)(x + 3) = 0$, so $x = 5$ (a length must be positive). The sides are $5, 12, 13$.

Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$, then cancel $(x - 3)$: $\dfrac{(x - 3)(x + 3)}{x - 3} = x + 3$.

Numerator $x^2 + 5x + 6 = (x + 2)(x + 3)$; denominator $x^2 + 4x + 4 = (x + 2)^2$. Cancel one $(x + 2)$: $\dfrac{(x + 2)(x + 3)}{(x + 2)^2} = \dfrac{x + 3}{x + 2}$. ($\dfrac{x + 2}{x + 3}$ inverted the result.)

$(x + 3)^2 = x^2 + 6x + 9$ and $(x - 3)^2 = x^2 - 6x + 9$, so $(x^2 + 6x + 9) - (x^2 - 6x + 9) = 12x$. (Or as a difference of squares with $a = x + 3$, $b = x - 3$: $(a + b)(a - b) = (2x)(6) = 12x$.) (0 assumed the two squares are equal.)

Compare with $(x + a)^2 = x^2 + 2ax + a^2$: $2ax = 12x$ gives $a = 6$, and $a^2 = 36$ confirms $a = 6$ (positive, since the middle term is $+12x$). ($a = 12$ used the middle coefficient without halving; $a = -6$ gives a $-12x$ middle term.)

Extract the common factor $x$: $x^3 + 5x^2 + 6x = x(x^2 + 5x + 6)$, then factor the trinomial: $x(x + 2)(x + 3)$. ($x(x^2 + 5x + 6)$ stopped before factoring the trinomial; $x(x + 6)(x + 1)$ uses numbers that sum to 7, not 5.)