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Adapted from Q1 of the sample paper for Unit 1

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a)  Using \"order of operations\" we first need to expand, or multiply out, the brackets.

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We have:

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\\( \\simplify{{C1} *(x+{C2})+{C3}} \\)

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so EVERYTHING inside the bracket must get multiplied by \\( \\var{C1}\\)

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\\( \\simplify{{C1} *(x+{C2})+{C3}} \\)

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becomes:

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\\( \\simplify{{C1}*x+{C1}*{C2}} \\simplify{{C3}}  \\)

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Finally, we can collect \"like terms\" to give:

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\\( \\simplify{{C1}*x+{C1}*{C2} +{C3}}  \\)

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b)  When an expression can be viewed as the difference of two perfect squares, i.e. \\(a^2-b^2\\), then we can factor it as \\((a+b)(a-b)\\).

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We have:

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\\(  x^2-\\var{Sq} \\)

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So this can be factorised as:

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\\(  (x+\\var{root})(x-\\var{root})   \\)

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Note: This relies on you being able to recognise square numbers.

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c)   Rearrange (Transpose):

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\\( R_1=R_0  (1+\\alpha t)       \\)

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\\( R_1=R_0 +R_0 \\alpha t       \\)

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\\( R_0 \\alpha t = R_0 - R_1      \\)

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\\(  \\alpha = \\frac{R_0 - R_1}{R_0 t }      \\)

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Constant #1

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Constant #2

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Constant #3

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List of square numbers to be used in \"difference of two squares\"

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The random choice of square number to use

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root of the chosen Sq

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Remove the brackets and simplify:

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\\( \\simplify{{C1} *(x+{C2})+{C3}} \\)

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Factorise the expression:

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\\(  x^2-\\var{Sq} \\)

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Transpose the formula to make \\( \\alpha  \\) the subject.

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\\( R_1=R_0  (1+\\alpha t)       \\)

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\\( \\alpha = \\) [[0]]

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This is algebraically correct but is a much clumsier transposition.

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