Factorise fully 27x^2 - 3

First, work out what number is a common factor of both terms - in other words what number 'goes into' both 27 and 3? 3 is a factor of both so put 3 outside the bracket. = 3( __ - __) Now work out what goes in each of the gaps inside the bracket - what do you need to multiply by 3 to get 27x2 ? What about to get - 3 ? = 3(9x2-1) But make sure you don't stop there! We need to factorise the expression inside the bracket, to make sure we have done what the exam question asked and factorised 'fully'. You may recognise that this expression looks familiar - it is the difference of 2 squares because there is no x term, only an x2 and a number. So we know the factorised version must look like this: =3(_x + _)(_x - _) To find the missing numbers we just need to work out what multiplies by itself to make 9 & what multiplies by itself to make 1. = 3(3x+1)(3x-1)

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Answered by Rebecca T. Maths tutor

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Given 6x+2y=4 and 5x+y=8, solve the simultaneous equations to find x and y.


A)Write x^2 – 8x + 25 in the form (x – a)^ 2 + b. (B) Write down the coordinates of the turning point of the graph of y = x2 – 8x + 25. (C)Hence describe the single transformation which maps the graph of y = x2 onto the graph of y = x2 – 8x + 25.


Solve 3x^2 = 8x - 2 giving your answers to 2 d.p.


Factorise and solve x^2 - 8x + 15 = 0.


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