factorise (x-3)^2 - 16?

[(x-3) + 4] [(x-3) -4]

[(x - 3) - 4] [ (x - 3) + 4 ]
(x - 7) (x + 1)

(x-3)^2 - 16 = (x-3-4)(x-3+4) = (x-7)(x+1)

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(x-16)(x-2) = 0

x^2-18x+32=0 x^2-16x-2x+32=0 x(x-16)-2(x-16)=0 (x-16)=0 (x-2)=0 x=16 x=2

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RE:
factorise (x-3)^2 - 16?
homework pl help

(x-3)^2 - 16?
(x^2 - 6x +9) - 16
x^2 -6x +9 - 16
x^2 -6x -7
(x-7)(x+1)

=(x-3-4)(x-3+4)
by using a^2-b^2=(a+b)(a-b)
=(x-7)(x+1)
=ans.

(x - 3)^2 - 16
= (x - 3)(x - 3) - 16
= x^2 - 3x - 3x + 9 - 16
= x^2 - 6x - 7
= (x + 1)(x - 7)

(x-3)^2 =16.
x-3= +or-4.
Taking +value x-3= 4. x=7.
Taking - value x-3= -4, x= -4+3= -1.
So the values of x= 7, -1.