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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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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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#### \$$\\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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