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This is the second of our many quizzes on Factoring. Click on this to access more such competitive quizzes on Algebra.

Factoring Quiz 2

mathematics

 Evaluate and factorize $\dfrac{5}{x} - \dfrac{1}{3} = \dfrac{1}{x}$ with the value of $x$.

 To factor, multiply the numerator and denominator term by Lowest Common Denominator and then evaluate the expression.

  The given expression we have: $ \\ \dfrac{5}{x} - \dfrac{1}{3} = \dfrac{1}{x} \\ $ Multiply the numerator and denominator term by the LCD i.e., Lowest Common Denominator as: $ \\  \Rightarrow 3x \cdot \left( {\dfrac{5}{x} - \dfrac{1}{3}} \right) = 3x \cdot \left( {\dfrac{1}{x}} \right) \\ \Rightarrow 3x \cdot \dfrac{5}{x} - 3x \cdot \dfrac{1}{3} = 3x \cdot \dfrac{1}{x} \\ \Rightarrow 15 - x = 3 \\ \Rightarrow x = 12$

 Factor out the terms $y\sqrt {98y}  - \sqrt {8{y^3}} $.

 Simplify the radicals and factor out the terms.

  Given, $y\sqrt {98y}  - \sqrt {8{y^3}} $ $ \\ $ Let us simplify the radicals as: $ \\ \Rightarrow y\sqrt {49 \cdot 2y}  - \sqrt {4 \cdot 2 \cdot {y^2} \cdot y}  \\  \Rightarrow y\sqrt {49}  \cdot \sqrt {2y}  - \sqrt 4  \cdot \sqrt {{y^2}}  \cdot \sqrt {2y} \\ \Rightarrow 7y\sqrt {2y}  - 2y\sqrt {2y} \\ \Rightarrow 5y\sqrt {2y} $

 Subtract and factor out the expression $\dfrac{{2x + 7}}{{\left( {x + 5} \right)\left( {x - 3} \right)}} - \dfrac{{x + 10}}{{\left( {x + 5} \right)\left( {x - 3} \right)}}$.

 To solve, factor out and simplify the numerator terms.

Given, $\dfrac{{2x + 7}}{{\left( {x + 5} \right)\left( {x - 3} \right)}} - \dfrac{{x + 10}}{{\left( {x + 5} \right)\left( {x - 3} \right)}}$ $ \\ $ Simplify the numerator terms, as there is a common denominator as: $ \\ \Rightarrow \dfrac{{\left( {2x + 7} \right) - \left( {x + 10} \right)}}{{\left( {x + 5} \right)\left( {x - 3} \right)}} \\ \Rightarrow \dfrac{{2x + 7 - x - 10}}{{\left( {x + 5} \right)\left( {x - 3} \right)}} \\ \Rightarrow \dfrac{{x - 3}}{{\left( {x + 5} \right)\left( {x - 3} \right)}} \\ \Rightarrow \dfrac{1}{{x + 5}}$

 Factorize $2a\sqrt {125{a^2}b}  - {a^2}\sqrt {80b}  + 4\sqrt {20{a^4}b} $

 Factor out the numerator terms and solve to get the required answer.

 Given, $2a\sqrt {125{a^2}b}  - {a^2}\sqrt {80b}  + 4\sqrt {20{a^4}b} $ $ \\ $ Simplifying the radicals, of the given expression we have: $ \\ \Rightarrow 2a\sqrt {25 \cdot 5 \cdot {a^2} \cdot b}  - {a^2}\sqrt {16 \cdot 5 \cdot b}  + 4\sqrt {4 \cdot 5 \cdot {a^4} \cdot b} \\ \Rightarrow 2a \cdot 5\sqrt {5 \cdot {a^2} \cdot b}  - {a^2} \cdot 4\sqrt {5 \cdot b}  + 4 \cdot 2\sqrt {5 \cdot {a^4} \cdot b} \\ \Rightarrow 2a \cdot 5 \cdot a\sqrt {5 \cdot b}  - {a^2} \cdot 4\sqrt {5 \cdot b}  + 4 \cdot 2 \cdot {a^2}\sqrt {5 \cdot b} \\ \Rightarrow 10{a^2}\sqrt {5b}  - 4{a^2}\sqrt {5b}  + 8{a^2}\sqrt {5b} \\ \Rightarrow 14{a^2}\sqrt {5b} $

 Factorise the expression $\dfrac{{{x^2} - 9x + 18}}{{{x^2} - 13x + 36}} - \dfrac{x}{{x - 4}}$.

 Factor the denominator term and then simplify it.

 Given, $\dfrac{{{x^2} - 9x + 18}}{{{x^2} - 13x + 36}} - \dfrac{x}{{x - 4}}$ $ \\ $ Factor the denominator term and the factorisation of  ${x^2} - 13x + 36$ is $\left( {x - 4} \right)\left( {x - 9} \right)$, hence we get: $ \\ \Rightarrow \dfrac{{{x^2} - 9x + 18}}{{\left( {x - 4} \right)\left( {x - 9} \right)}} - \dfrac{x}{{x - 4}} \\ \Rightarrow \dfrac{{{x^2} - 9x + 18}}{{\left( {x - 4} \right)\left( {x - 9} \right)}} - \dfrac{x}{{x - 4}} \cdot \dfrac{{x - 9}}{{x - 9}} \\ \Rightarrow \dfrac{{{x^2} - 9x + 18}}{{\left( {x - 4} \right)\left( {x - 9} \right)}} - \dfrac{{x\left( {x - 9} \right)}}{{\left( {x - 4} \right)\left( {x - 9} \right)}} \\ \Rightarrow \dfrac{{{x^2} - 9x + 18 - x\left( {x - 9} \right)}}{{\left( {x - 4} \right)\left( {x - 9} \right)}} \\  \Rightarrow \dfrac{{18}}{{\left( {x - 4} \right)\left( {x - 9} \right)}}$