X^212yy^2 Factorise Get the answers you need, now!Factorise x^(3)2x^(2)x2 Factorise x^(3)2x^(2)x The locus of the midpoint of a chord of the circle x^2 y^2 2x 2y 23=0, of length 8 units is (A) x^2 y^2 x y 1 =0 (B) x^2 y^2 2x 2y 7 = 0 x^2 y^2 2x 2y 1 = 0 (D) x^2 y^2 2x 2y 5 = 0Bo Cephas Answered 2 years ago Author has 144 answers and 1197K answer views Rearrange the terms as follows x^2 — y^2 — 2y — 1 Group the last three terms, factoring out a —1 x^2 — (y^2 2y 1) Those last three terms are a perfect square x^2 — (y 1)^2 Now you have the difference of two squares
Factorise X 5 Y 5
