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
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Explanation 1 = xy −(x2 − y2)2y Rewrite as difference of two terms Rewrite the terms as squares The difference of squares can be factored using the rule a^ {2}b^ {2}=\left (aFactorisex^212yy^2 186 results algebra factorise 3px3pyxy Math Factorise X'29X 14 Maths factorise Factorise 3px3pyxy Business Math a) Solve for x 2x² 5x = 3 b) Solve for x and y x y = 4 and x² y² xy 12 =0 c) Solve for x 8 2x 3 = ½x² d) Factorise 6x² 7xy 3y² e) Factorise 4x³ 5x²y 4xy²Factor 2y^2y1 2y2 − y − 1 2 y 2 y 1 For a polynomial of the form ax2 bx c a x 2 b x c, rewrite the middle term as a sum of two terms whose product is a⋅c = 2⋅−1 = −2 a ⋅ c = 2 ⋅ 1 = 2 and whose sum is b = −1 b = 1 Tap for more steps Factor − 1 1 out of − y y
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