Advanced Math

QUESTION 1 [2 marks] Easy

Simplify: $3(2x+4)-5x$.

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Distribute: $6x+12-5x=x+12$.

QUESTION 2 [2 marks] Easy

Simplify: $\dfrac{x^2\cdot x^5}{x^3}$ (for $x\neq0$).

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Using exponent rules: $\dfrac{x^{2+5}}{x^3}=\dfrac{x^7}{x^3}=x^{7-3}=x^4$.

QUESTION 3 [3 marks] Medium

Expand and simplify: $(x+3)(x-3)$.

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This is a difference of squares: $(x+3)(x-3)=x^2-3x+3x-9=x^2-9$.

QUESTION 4 [3 marks] Medium

Simplify: $\dfrac{4x^2-9}{2x+3}$ (for $x\neq-\tfrac{3}{2}$).

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Factor the numerator as a difference of squares: $4x^2-9=(2x-3)(2x+3)$. So $\dfrac{(2x-3)(2x+3)}{2x+3}=2x-3$.

QUESTION 5 [4 marks] Hard

Simplify fully: $\dfrac{x^2+5x+6}{x^2-4}\cdot\dfrac{x-2}{x+3}$ (for $x\neq2,-2,-3$).

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Factor: $x^2+5x+6=(x+2)(x+3)$ and $x^2-4=(x-2)(x+2)$. So the product is $\dfrac{(x+2)(x+3)}{(x-2)(x+2)}\cdot\dfrac{x-2}{x+3}=\dfrac{(x+2)(x+3)(x-2)}{(x-2)(x+2)(x+3)}=1$, after cancelling the common factors $(x+2)$, $(x+3)$, and $(x-2)$.

QUESTION 6 [5 marks] Hard

What is the positive value of $x$ for which the expression $\dfrac{x+4}{x^2-x-12}$ is undefined?

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The expression is undefined where the denominator is 0. Factor: $x^2-x-12=(x-4)(x+3)$, so the denominator is 0 when $x=4$ or $x=-3$. The positive value is $x=4$.

QUESTION 7 [5 marks] Hard

Write $\dfrac{1}{x-2}+\dfrac{1}{x+2}$ as a single fraction (for $x\neq\pm2$).

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Using the common denominator $(x-2)(x+2)=x^2-4$: $\dfrac{1}{x-2}+\dfrac{1}{x+2}=\dfrac{(x+2)+(x-2)}{(x-2)(x+2)}=\dfrac{2x}{x^2-4}$.

Equivalent expressions 7 questions