Factorise the following by taking out the common factor
6xy – 4y2 + 12xy – 2yzx​

2 thoughts on “Factorise the following by taking out the common factor<br />6xy – 4y2 + 12xy – 2yzx​”

1. 6xy + 12xy – 4y2 – 2yzx [Addition is commutative]

   = = (6 × x × y) + (2 × 6 × x × y) + (−1)(2)(2) y +

   y) + ((-1)(2)(y)(z)(x))

   Taking out 6 x x x y from first two terms and (-1) × 2 × y from last two terms we

   get

   = 6 x × x × y(1 + 2) + (−1)(2)y [2y+zx]

   = 6 xy(3) = 2y(2y+zx)

   = (2 × 3 ×3 × x × y) – 2xy(2y + zx)

   Taking out 2y from two terms

   = 2y(9x – (2y + zx))

   = 2y(9x – 2y – xz)

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2. ❥Δnsɯer࿐

   6xy + 12xy – 4y2 – 2yzx

   [∵ Addition is commutative]

   = (6 × x × y) + (2 × 6 × x × y) + (–1)(2)(2) y + y) + ((–1)(2)(y)(z)(x))

   Taking out 6 x x x y from first two terms and (–1) × 2 × y from last two terms we get

   = 6 × x × y(1 + 2) + (–1)(2)y [2y + zx]

   = 6 × y(3) – 2y(2y + zx)

   = (2 × 3 × 3 × x × y) – 2xy(2y + zx)

   Taking out 2y from two terms

   = 2y(9x – (2y + zx))

   = 2y(9x – 2y – xz)

   Reply