Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

Q1) \({x^2 +8x+15}\over{x+5}\) = [ \(x+3\) ]

Q1) \({x^2 -9}\over{x+3}\) = [ \(x-3\) ]

Q1) \({3x^2 +8x+4}\over{x+2}\) = [ \(3x+2\) ]

Q2) \({x-3\over{x^2 -7x+12}}\) = [ \(1\over{x-4}\) ]

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

Q2) \({4x^2 -21x-18}\over{x-6}\) = [ \(4x+3\) ]

Q3) \({x^2 +2x-24}\over{x-4}\) = [ \(x+6\) ]

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

Q3) \({4x^2 +18x-36}\over{x+6}\) = [ \(4x-6\) ]

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

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

Q4) \({3x^2 +x-10}\over{x+2}\) = [ \(3x-5\) ]

Q5) \({x+2\over{x^2 -2x-8}}\) = [ \(1\over{x-4}\) ]

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

Q5) \({4x^2 +13x+10}\over{x+2}\) = [ \(4x+5\) ]

Q6) \({x^2 -7x+10}\over{x-2}\) = [ \(x-5\) ]

Q6) \({x-4}\over{x^2 -16}\) = [ \(1\over{x+4}\) ]

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

Q7) \({x-3\over{x^2 -x-6}}\) = [ \(1\over{x+2}\) ]

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

Q7) \({4x^2 +14x-8}\over{x+4}\) = [ \(4x-2\) ]

Q8) \({x-4\over{x^2 -6x+8}}\) = [ \(1\over{x-2}\) ]

Q8) \({x^2 -36}\over{x-6}\) = [ \(x+6\) ]

Q8) \({2x^2 +14x+12}\over{x+6}\) = [ \(2x+2\) ]

Q9) \({x-5\over{x^2 +x-30}}\) = [ \(1\over{x+6}\) ]

Q9) \({x+7}\over{x^2 -49}\) = [ \(1\over{x-7}\) ]

Q9) \({3x^2 -9x+6}\over{x-2}\) = [ \(3x-3\) ]

Q10) \({x^2 -9x+14}\over{x-2}\) = [ \(x-7\) ]

Q10) \({x^2 -25}\over{x-5}\) = [ \(x+5\) ]

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