Question

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

Answer 1

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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)

Answer 2

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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)