Algebraic techniques

Fluency

$4x(2x - 3)$ .
$5ab(a + 2b - 3)$ .
$\dfrac{3x^2}{y^3}$ .
$(3a^2)^3 = 27a^6$ . Then $27a^6 \times 2a^{-4} = 54a^2$ .
$x^2 - 3x - 28$ .
$4x^2 - 12x + 9$ .
$(x - 7)(x + 7)$ .
$(x + 6)(x - 3)$ .
LCD $= 2x$ : $\dfrac{8}{2x} + \dfrac{3}{2x} = \dfrac{11}{2x}$ .
When $u = 0$ : $s = \dfrac{1}{2}at^2$ , so $t^2 = \dfrac{2s}{a}$ , $t = \sqrt{\dfrac{2s}{a}}$ (taking the positive root).

Reasoning

Product $= 3 \times (-8) = -24$ . Numbers: $12$ and $-2$ . Split: $3x^2 + 12x - 2x - 8 = 3x(x + 4) - 2(x + 4) = (x + 4)(3x - 2)$ .
$x^2 - 6x + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8$ .
$\dfrac{(x-2)(x+2)}{(x+2)(x+3)} = \dfrac{x - 2}{x + 3}$ (for $x \neq -2$ ).
LCD $= (x-1)(x+2)$ : $\dfrac{2(x+2) - 3(x-1)}{(x-1)(x+2)} = \dfrac{2x + 4 - 3x + 3}{(x-1)(x+2)} = \dfrac{-x + 7}{(x-1)(x+2)}$ .
$2E = mv^2$ , $v^2 = \dfrac{2E}{m}$ , $v = \sqrt{\dfrac{2E}{m}}$ .
$y(x - 2) = 3x + 1$ , $xy - 2y = 3x + 1$ , $xy - 3x = 2y + 1$ , $x(y - 3) = 2y + 1$ , $x = \dfrac{2y + 1}{y - 3}$ .
$\dfrac{(x+3)^2}{(x-3)(x+3)} = \dfrac{x + 3}{x - 3}$ (for $x \neq -3$ ).
$2(x^2 - 9) = 2(x - 3)(x + 3)$ .

Reasoning

$x^2 - 4 = (x-2)(x+2)$ uses the difference of two squares. For $x^2 + 4$ , there are no real numbers $a, b$ with $a^2 - b^2 = x^2 + 4$ in the required form. Since $x^2 \geq 0$ , $x^2 + 4 \geq 4 > 0$ , so it never equals zero and cannot be split into real linear factors.
$x^2 + 6x + 11 = (x + 3)^2 - 9 + 11 = (x + 3)^2 + 2$ . Since $(x+3)^2 \geq 0$ , the expression $\geq 2 > 0$ for all real $x$ .
LCD $= (x+1)(x+2)(x+3)$ . Numerator: $(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)$ $= (x^2 + 5x + 6) + (x^2 + 4x + 3) + (x^2 + 3x + 2)$ $= 3x^2 + 12x + 11$ . Result: $\dfrac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)}$ .
$S = 2\pi r(r + h)$ . Rearranging: $r + h = \dfrac{S}{2\pi r}$ , so $h = \dfrac{S}{2\pi r} - r = \dfrac{S - 2\pi r^2}{2\pi r}$ .
$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ .

Reasoning

Product $= 6 \times (-3) = -18$ . Numbers: $-9$ and $2$ . Split: $6x^2 - 9xy + 2xy - 3y^2 = 3x(2x - 3y) + y(2x - 3y) = (2x - 3y)(3x + y)$ . Check: $(2x - 3y)(3x + y) = 6x^2 + 2xy - 9xy - 3y^2 = 6x^2 - 7xy - 3y^2$ . Correct.
$\left(a + \dfrac{1}{a}\right)^2 = a^2 + 2 + \dfrac{1}{a^2}$ . So $25 = a^2 + 2 + \dfrac{1}{a^2}$ , hence $a^2 + \dfrac{1}{a^2} = 23$ .
$\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{(n+1) - n}{n(n+1)} = \dfrac{1}{n(n+1)}$ . The sum telescopes: $\left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{99} - \dfrac{1}{100}\right) = 1 - \dfrac{1}{100} = \dfrac{99}{100}$ .
Width $= \dfrac{x^2 + 7x + 10}{x + 5} = \dfrac{(x+5)(x+2)}{x+5} = x + 2$ . Perimeter $= 2(x + 5 + x + 2) = 2(2x + 7) = 4x + 14$ . If perimeter $= 26$ : $4x + 14 = 26$ , $x = 3$ .

Year 10 core - answers