Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

Q1) \({x+2\over{x^2 -3x-10}}\) = [ \(1\over{x-5}\) ]

Q1) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q1) \({3x^2 -12x-36}\over{x-6}\) = [ \(3x+6\) ]

Q2) \({x+3\over{x^2 -x-12}}\) = [ \(1\over{x-4}\) ]

Q2) \({x+5}\over{x^2 -25}\) = [ \(1\over{x-5}\) ]

Q2) \({2x^2 +4x-30}\over{x+5}\) = [ \(2x-6\) ]

Q3) \({x-3\over{x^2 -x-6}}\) = [ \(1\over{x+2}\) ]

Q3) \({x-2}\over{x^2 -4}\) = [ \(1\over{x+2}\) ]

Q3) \({4x^2 -18x-10}\over{x-5}\) = [ \(4x+2\) ]

Q4) \({x+5\over{x^2 -25}}\) = [ \(1\over{x-5}\) ]

Q4) \({x-4}\over{x^2 -16}\) = [ \(1\over{x+4}\) ]

Q4) \({5x^2 -31x+30}\over{x-5}\) = [ \(5x-6\) ]

Q5) \({x^2 +7x+10}\over{x+2}\) = [ \(x+5\) ]

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

Q5) \({2x^2 -x-10}\over{x+2}\) = [ \(2x-5\) ]

Q6) \({x-8\over{x^2 -x-56}}\) = [ \(1\over{x+7}\) ]

Q6) \({x^2 -16}\over{x-4}\) = [ \(x+4\) ]

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

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

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

Q7) \({5x^2 +18x+9}\over{x+3}\) = [ \(5x+3\) ]

Q8) \({x-2\over{x^2 -4x+4}}\) = [ \(1\over{x-2}\) ]

Q8) \({x^2 -16}\over{x+4}\) = [ \(x-4\) ]

Q8) \({5x^2 +19x+12}\over{x+3}\) = [ \(5x+4\) ]

Q9) \({x-2\over{x^2 -6x+8}}\) = [ \(1\over{x-4}\) ]

Q9) \({x-3}\over{x^2 -9}\) = [ \(1\over{x+3}\) ]

Q9) \({2x^2 -15x+18}\over{x-6}\) = [ \(2x-3\) ]

Q10) \({x^2 -2x-8}\over{x+2}\) = [ \(x-4\) ]

Q10) \({x^2 -9}\over{x-3}\) = [ \(x+3\) ]

Q10) \({5x^2 +22x+8}\over{x+4}\) = [ \(5x+2\) ]