Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

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

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

Q1) \({2x^2 -14x+12}\over{x-6}\) = [ \(2x-2\) ]

Q2) \({x^2 -6x-27}\over{x+3}\) = [ \(x-9\) ]

Q2) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q2) \({3x^2 +9x-12}\over{x+4}\) = [ \(3x-3\) ]

Q3) \({x-7\over{x^2 -13x+42}}\) = [ \(1\over{x-6}\) ]

Q3) \({x-3}\over{x^2 -9}\) = [ \(1\over{x+3}\) ]

Q3) \({2x^2 -9x+9}\over{x-3}\) = [ \(2x-3\) ]

Q4) \({x-5\over{x^2 -7x+10}}\) = [ \(1\over{x-2}\) ]

Q4) \({x-8}\over{x^2 -64}\) = [ \(1\over{x+8}\) ]

Q4) \({5x^2 -12x+4}\over{x-2}\) = [ \(5x-2\) ]

Q5) \({x+3\over{x^2 +8x+15}}\) = [ \(1\over{x+5}\) ]

Q5) \({x-9}\over{x^2 -81}\) = [ \(1\over{x+9}\) ]

Q5) \({3x^2 +16x-12}\over{x+6}\) = [ \(3x-2\) ]

Q6) \({x-3\over{x^2 -10x+21}}\) = [ \(1\over{x-7}\) ]

Q6) \({x+3}\over{x^2 -9}\) = [ \(1\over{x-3}\) ]

Q6) \({3x^2 +7x-6}\over{x+3}\) = [ \(3x-2\) ]

Q7) \({x+6\over{x^2 +11x+30}}\) = [ \(1\over{x+5}\) ]

Q7) \({x+2}\over{x^2 -4}\) = [ \(1\over{x-2}\) ]

Q7) \({3x^2 -12x-36}\over{x-6}\) = [ \(3x+6\) ]

Q8) \({x^2 -2x-8}\over{x-4}\) = [ \(x+2\) ]

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

Q8) \({5x^2 +24x+16}\over{x+4}\) = [ \(5x+4\) ]

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

Q9) \({x^2 -81}\over{x+9}\) = [ \(x-9\) ]

Q9) \({3x^2 -15x+12}\over{x-4}\) = [ \(3x-3\) ]

Q10) \({x-3\over{x^2 -x-6}}\) = [ \(1\over{x+2}\) ]

Q10) \({x^2 -16}\over{x+4}\) = [ \(x-4\) ]

Q10) \({4x^2 +22x+10}\over{x+5}\) = [ \(4x+2\) ]