Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

Q1) \({x+2\over{x^2 -2x-8}}\) = [ \(1\over{x-4}\) ]

Q1) \({x+2}\over{x^2 -4}\) = [ \(1\over{x-2}\) ]

Q1) \({3x^2 -12}\over{x+2}\) = [ \(3x-6\) ]

Q2) \({x+2\over{x^2 +11x+18}}\) = [ \(1\over{x+9}\) ]

Q2) \({x^2 -25}\over{x-5}\) = [ \(x+5\) ]

Q2) \({2x^2 +14x+12}\over{x+6}\) = [ \(2x+2\) ]

Q3) \({x+2\over{x^2 -5x-14}}\) = [ \(1\over{x-7}\) ]

Q3) \({x+5}\over{x^2 -25}\) = [ \(1\over{x-5}\) ]

Q3) \({5x^2 -28x+15}\over{x-5}\) = [ \(5x-3\) ]

Q4) \({x+9\over{x^2 +6x-27}}\) = [ \(1\over{x-3}\) ]

Q4) \({x^2 -9}\over{x+3}\) = [ \(x-3\) ]

Q4) \({2x^2 +15x+25}\over{x+5}\) = [ \(2x+5\) ]

Q5) \({x-7\over{x^2 -9x+14}}\) = [ \(1\over{x-2}\) ]

Q5) \({x-4}\over{x^2 -16}\) = [ \(1\over{x+4}\) ]

Q5) \({3x^2 +17x+20}\over{x+4}\) = [ \(3x+5\) ]

Q6) \({x+10\over{x^2 +6x-40}}\) = [ \(1\over{x-4}\) ]

Q6) \({x-2}\over{x^2 -4}\) = [ \(1\over{x+2}\) ]

Q6) \({2x^2 -x-6}\over{x-2}\) = [ \(2x+3\) ]

Q7) \({x^2 +x-12}\over{x-3}\) = [ \(x+4\) ]

Q7) \({x^2 -4}\over{x+2}\) = [ \(x-2\) ]

Q7) \({5x^2 +13x-6}\over{x+3}\) = [ \(5x-2\) ]

Q8) \({x^2 -5x+6}\over{x-3}\) = [ \(x-2\) ]

Q8) \({x+3}\over{x^2 -9}\) = [ \(1\over{x-3}\) ]

Q8) \({2x^2 +12x+18}\over{x+3}\) = [ \(2x+6\) ]

Q9) \({x+6\over{x^2 -x-42}}\) = [ \(1\over{x-7}\) ]

Q9) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q9) \({3x^2 +11x+10}\over{x+2}\) = [ \(3x+5\) ]

Q10) \({x+2\over{x^2 +7x+10}}\) = [ \(1\over{x+5}\) ]

Q10) \({x^2 -36}\over{x+6}\) = [ \(x-6\) ]

Q10) \({5x^2 +20x+15}\over{x+3}\) = [ \(5x+5\) ]