By factorising, sketch the graph $y = x^2 - 9x + 18$, showing the points of intersection with the coordinate axes. [2] $(x-3)(x-6)$, intercepts: $x=3,6$, $y$-intercept: $(0,18)$
By factorising, sketch the graph $y = x^2 + 3x - 28$, showing the points of intersection with the coordinate axes. [2] $(x+7)(x-4)$, intercepts: $x=-7,4$, $y$-intercept: $(0,-28)$
By factorising, sketch the graph $y = x^2 - 16x + 64$, showing the points of intersection with the coordinate axes. [2] $(x-8)^2$, intercepts: $x=8$ (double root), $y$-intercept: $(0,64)$
By factorising, sketch the graph $y = 4x + x^2 - 45$, showing the points of intersection with the coordinate axes. [2] $(x+9)(x-5)$, intercepts: $x=-9,5$, $y$-intercept: $(0,-45)$
By factorising, sketch the graph $y = x^2 - 49$, showing the points of intersection with the coordinate axes. [2] $(x+7)(x-7)$, intercepts: $x=-7,7$, $y$-intercept: $(0,-49)$
By factorising, sketch the graph $y = 4x^2 - 25$, showing the points of intersection with the coordinate axes. [2] $(2x+5)(2x-5)$, intercepts: $x=-\tfrac{5}{2}, \tfrac{5}{2}$, $y$-intercept: $(0,-25)$
By factorising, sketch the graph $y = 3x^2 + 7x + 2$, showing the points of intersection with the coordinate axes. [2] $(3x+1)(x+2)$, intercepts: $x=-2,-\tfrac{1}{3}$, $y$-intercept: $(0,2)$
By factorising, sketch the graph $y = 5 + 9x + 4x^2$, showing the points of intersection with the coordinate axes. [2] $(4x+5)(x+1)$, intercepts: $x=-\tfrac{5}{4},-1$, $y$-intercept: $(0,5)$
By factorising, sketch the graph $y = -x^2 + 3x + 18$, showing the points of intersection with the coordinate axes. [2] $-(x-6)(x+3)$, intercepts: $x=-3, 6$, $y$-intercept: $(0,18)$
By factorising, sketch the graph $y = -x^2 - 7x + 8$, showing the points of intersection with the coordinate axes. [2] $-(x+8)(x-1)$, intercepts: $x=-8, 1$, $y$-intercept: $(0,8)$
By factorising, sketch the graph $y = -3x^2 + 10x + 8$, showing the points of intersection with the coordinate axes. [2] $-(3x+2)(x-4)$, intercepts: $x=-\tfrac{2}{3}, 4$, $y$-intercept: $(0,8)$
By factorising, sketch the graph $y = -2x^2 - 4x + 30$, showing the points of intersection with the coordinate axes. [2] $-2(x+5)(x-3)$, intercepts: $x=-5, 3$, $y$-intercept: $(0,30)$
Solve $2x^2 - 11x + 12 = 0$. [2] $x = \tfrac{3}{2},\; 4$
Solve $6 - x - x^2 = 0$. [2] $x = -3,\; 2$
Solve $5x^2 + 13x - 6 = 0$. [2] $x = -3,\; \tfrac{2}{5}$
Solve $3 - 14x - 5x^2 = 0$. [2] $x = -3,\; \tfrac{1}{5}$
Solve $(x+4)(x-1)(x-6) = 0$. [1] $x = -4,\; 1,\; 6$
Solve $x^2(x+3)(x-5) = 0$. [1] $x = -3,\; 0,\; 5$
Solve $x - 8 + \dfrac{12}{x} = 0$. [2] $x = 2,\; 6$
Solve $\dfrac{x+10}{x+4} = x$. [2] $x = -5,\; 2$
Solve $(x-4)(x+2) = 7$. [2] $x = -3,\; 5$
Solve $x^3 + 3x^2 - 18x = 0$. [2] $x = -6,\; 0,\; 3$
Solve $3x^4 - 10x^3 + 8x^2 = 0$. [2] $x = 0,\; \tfrac{4}{3},\; 2$
Solve $4x^3 + 4x^2 - 3x = 0$. [2] $x = -\tfrac{3}{2},\; 0,\; \tfrac{1}{2}$
By factorising, simplify $$\dfrac{x^2 - 9}{x^2 + 5x + 6}$$ [2] $\dfrac{x-3}{x+2}$
By factorising, simplify $$\dfrac{x^2 - 7x + 12}{x^2 - 16}$$ [2] $\dfrac{x-3}{x+4}$
By factorising, simplify $$\dfrac{2x^2 + 5x - 3}{4x^2 - 1}$$ [2] $\dfrac{x+3}{2x+1}$
By factorising, simplify $$\dfrac{x^3 - 4x}{x^2 + 5x + 6}$$ [2] $\dfrac{x(x-2)}{x+3}$
Fully factorise $x^4 - 7x^2 + 10$ [2] $(x^2-2)(x^2-5)$
Fully factorise $x^6 + 9x^3 + 20$ [2] $(x^3+4)(x^3+5)$
Fully factorise $9x^4 - 25$ [1] $(3x^2-5)(3x^2+5)$
Fully factorise $3x^4 + 10x^2 + 8$ [2] $(3x^2+4)(x^2+2)$
Fully factorise $2x^5 - 7x^3 - 4x$ [2] $x(2x^2+1)(x+2)(x-2)$
Fully factorise $4x^6 - 8x^4 - 12x^2$ [2] $4x^2(x^2-3)(x^2+1)$
Factorise $x^2 - 5$ into two linear factors with real coefficients. [1] $(x-\sqrt{5})(x+\sqrt{5})$
Factorise $x^2 - 12$ into two linear factors with real coefficients, simplifying any surds. [2] $(x-2\sqrt{3})(x+2\sqrt{3})$
Factorise $x^2 - 7$ and hence solve $x^2 - 7 = 0$, giving your answers exactly. [2] $(x-\sqrt{7})(x+\sqrt{7}) \Rightarrow x = \pm\sqrt{7}$
Fully factorise $x^4 - 9$ over the real numbers. [2] $(x^2+3)(x-\sqrt{3})(x+\sqrt{3})$
Factorise $x^{\frac{3}{2}} - 4x^{\frac{1}{2}}$ [1] $x^{\frac{1}{2}}(x - 4)$
Factorise $2x^{\frac{5}{2}} + 6x^{\frac{3}{2}}$ [1] $2x^{\frac{3}{2}}(x + 3)$
By factorising, solve $x^{\frac{3}{2}} = 9x^{\frac{1}{2}}$ [2] $x^{\frac{1}{2}}(x - 9) = 0 \Rightarrow x = 0, 9$
Factorise $3x^{-1} + 6x^{-2}$ [1] $3x^{-2}(x + 2)$
Factorise and solve $x^4 - 13x^2 + 36 = 0$ [2] $(x^2 - 4)(x^2 - 9) \Rightarrow x = \pm 2, \pm 3$
Factorise and solve $-x^4 + 5x^2 - 4 = 0$ [2] $(x^2 - 1)(4 - x^2) \Rightarrow x = \pm 1, \pm 2$
Factorise and solve $(x-2)^2 - 5(x-2) + 6 = 0$ [2] $(x-2 - 2)(x-2 - 3) \Rightarrow x = 4, 5$
Factorise and solve $2^{2x} - 8\times 2^x = 0$ [2] $2^x(2^x - 8) \Rightarrow x = 3$
Factorise and solve $5^{2x} - 6\times 5^x + 5 = 0$ [2] $(5^x - 1)(5^x - 5) \Rightarrow x = 0, 1$
Factorise and solve $x - 7\sqrt{x} + 12 = 0$ [2] $(\sqrt{x} - 3)(\sqrt{x} - 4) \Rightarrow x = 9, 16$
Factorise and solve $9^x - 4\times 3^x + 3 = 0$ [2] $(3^x - 1)(3^x - 3) \Rightarrow x = 0, 1$
Factorise and solve $4^x - 3\times 2^{x+1} + 8 = 0$ [2] $(2^x - 2)(2^x - 4) \Rightarrow x = 1, 2$
Factorise and solve $3^{2x+1} - 10\times 3^x + 3 = 0$ [2] $(3\times 3^x - 1)(3^x - 3) \Rightarrow x = -1, 1$
Factorise and solve $2^{2x+1} - 5\times 2^x + 2 = 0$ [2] $(2\times 2^x - 1)(2^x - 2) \Rightarrow x = -1, 1$
Factorise $x^4 + 4$ into two quadratic factors with integer coefficients. [2] $(x^2 + 2x + 2)(x^2 - 2x + 2)$
Show that $n^4 + 4$ is never prime for integers $n > 1$. [2] $n^4 + 4 = (n^2+2n+2)(n^2-2n+2)$. For $n > 1$, $n^2 - 2n + 2 = (n-1)^2 + 1 \geq 2$ and $n^2 + 2n + 2 > 2$, so $n^4 + 4$ is a product of two factors both greater than $1$, hence not prime.
Factorise $x^4 + x^2 + 1$ into two quadratic factors with integer coefficients. [2] $(x^2 + x + 1)(x^2 - x + 1)$
Factorise $x^4 + 1$ into two quadratic factors with real coefficients. [2] $(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$
Fully factorise $x^4 + 5x^3 + 8x^2 + 5x + 1$
Hint: divide by $x^2$ and consider $t = x + \dfrac{1}{x}$. [3] $(x+1)^2(x^2+3x+1)$
You are given the identity $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ Given that $x^3 + y^3 + z^3 = 3xyz$ and $x + y + z \neq 0$, prove that $x = y = z$. [3] Since $x+y+z \neq 0$, the identity forces $x^2+y^2+z^2-xy-yz-zx = 0$. This equals $\frac{1}{2}\left[(x-y)^2 + (y-z)^2 + (z-x)^2\right]$, a sum of squares, so each square is zero and $x = y = z$.
Factorise $x^5 + x + 1$ into a quadratic factor and a cubic factor with integer coefficients. [2] $(x^2 + x + 1)(x^3 - x^2 + 1)$
By factorising, find all positive integer solutions of $$x^3 + 3x^2y + 3xy^2 + y^3 - x - y = 24$$ [3] $(x+y)^3 - (x+y) = 24$, so $x+y = 3$, giving $(x, y) = (1, 2)$ or $(2, 1)$.