Mr Daniels Maths
Difference of two squares

Set 1

Set 2

Set 3

Q1) Expand \((x+10)(x-10)\) = [ \(x^2 - 100\)]

Q1) Factorise \(x^2 - 361\)= [ \((x+19)(x-19)\)]

Q1) Factorise \(64x^2-49\)= [ \((8x + 7)(8x - 7)\)]

Q2) Expand \((x+1)(x-1)\) = [ \(x^2 - 1\)]

Q2) Factorise \(x^2 - 400\)= [ \((x+20)(x-20)\)]

Q2) Factorise \(64x^2-1\)= [ \((8x + 1)(8x - 1)\)]

Q3) Expand \((x+9)(x-9)\) = [ \(x^2 - 81\)]

Q3) Factorise \(x^2 - 49\)= [ \((x+7)(x-7)\)]

Q3) Factorise \(16x^2-16\)= [ \((4x + 4)(4x - 4)\)]

Q4) Expand \((x+6)(x-6)\) = [ \(x^2 - 36\)]

Q4) Factorise \(x^2 - 196\)= [ \((x+14)(x-14)\)]

Q4) Factorise \(25x^2-25\)= [ \((5x + 5)(5x - 5)\)]

Q5) Expand \((x+7)(x-7)\) = [ \(x^2 - 49\)]

Q5) Factorise \(x^2 - 324\)= [ \((x+18)(x-18)\)]

Q5) Factorise \(9x^2-1\)= [ \((3x + 1)(3x - 1)\)]

Q6) Expand \((x+9)(x-9)\) = [ \(x^2 - 81\)]

Q6) Factorise \(x^2 - 1\)= [ \((x+1)(x-1)\)]

Q6) Factorise \(9x^2-9\)= [ \((3x + 3)(3x - 3)\)]

Q7) Expand \((x+6)(x-6)\) = [ \(x^2 - 36\)]

Q7) Factorise \(x^2 - 36\)= [ \((x+6)(x-6)\)]

Q7) Factorise \(81x^2-4\)= [ \((9x + 2)(9x - 2)\)]

Q8) Expand \((x+6)(x-6)\) = [ \(x^2 - 36\)]

Q8) Factorise \(x^2 - 324\)= [ \((x+18)(x-18)\)]

Q8) Factorise \(4x^2-64\)= [ \((2x + 8)(2x - 8)\)]

Q9) Expand \((x+5)(x-5)\) = [ \(x^2 - 25\)]

Q9) Factorise \(x^2 - 324\)= [ \((x+18)(x-18)\)]

Q9) Factorise \(64x^2-1\)= [ \((8x + 1)(8x - 1)\)]

Q10) Expand \((x+10)(x-10)\) = [ \(x^2 - 100\)]

Q10) Factorise \(x^2 - 49\)= [ \((x+7)(x-7)\)]

Q10) Factorise \(36x^2-100\)= [ \((6x + 10)(6x - 10)\)]