You probably saw these in standard A-Level Maths, but in Further Maths, the denominators get messier. Linear Factors:
Her pen flew. She set up the identity: ( 5x^2 + 4x - 11 \equiv A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) ). She chose the cover-up rule for speed: ( x=1 ) gave ( A = 1 ). ( x=-2 ) gave ( B = -1 ). ( x=3 ) gave ( C = 5 ). core pure -as year 1- unit test 5 algebra and functions
your roots, even if the question asks for "show that" working. Substitution is your friend: You probably saw these in standard A-Level Maths,
: Forming new equations where the roots are modified (e.g., roots are ) without solving the original equation. She chose the cover-up rule for speed: (
She factorised: let ( u = x^2 ). Then ( u^2 - 8u + 16 = (u-4)^2 ). So ( p(x) = (x^2 - 4)^2 = (x-2)^2 (x+2)^2 ).
The first hurdle in any AS algebra test is the fluency of manipulation. The exam assumes you are comfortable with GCSE concepts and builds upon them with and surds .
Let $y = \frac1x$. Therefore $x = \frac1y$. Step 2: Substitute into the original equation: $$2\left(\frac1y\right)^3 - 3\left(\frac1y\right)^2 + 4\left(\frac1y\right) - 1 = 0$$ Step 3: Multiply through by $y^3$ to clear denominators: $$2 - 3y + 4y^2 - y^3 = 0$$ Step 4: Rearrange into standard polynomial form (highest power first): $$-y^3 + 4y^2 - 3y + 2 = 0$$ Or multiply by -1: $$y^3 - 4y^2 + 3y - 2 = 0$$