Question

(z - 1)^2 - 169 factorise by using identity​

Answer 1

Answer:

let a=(z-1)

let b=13

a^2=(z-1)^2

b^2=13^2=169

a^2-b^2=(a+b)*(a-b)

now put the value

(z-1+13)(z-1-13)

(z+12)(z-14)=(z-1)^2-169

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