Question

factorise:a^2+b^2-2ba+2bc-2ca​

Answer 1

Answer:

${a}^{2} + {b}^{2} - 2ba + 2bc - 2ca \\ = ( {a}^{2} + {b}^{2} - 2ba) + 2c(b - a) \\ = (b - a)^{2} - 2c(b - a) \\ = (b - a) \{(b - a) - 2c \}\\ = (b - a) \{b - a- 2c \}$

Answer 2

Answer:

a

2

+b

2

−2ba+2bc−2ca

=(a

2

+b

2

−2ba)+2c(b−a)

=(b−a)

2

−2c(b−a)

=(b−a){(b−a)−2c}

=(b−a){b−a−2c}