How do you factor #16-(y-3)^2#?

1 Answer
Jul 12, 2016

Answer:

(7 -y)(1 +y)

Explanation:

This expression is a #color(blue)"difference of squares"# which, in general factorises as.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|))) ........ (A)#

Now #16=4^2rArra=4" and " (y-3)^2rArrb=y-3#

Substitute these values for a and b into (A)

#rArr16-(y-3)^2=(4-(y-3))(4+y-3)#

Simplifying the brackets gives.

#(4-y+3)(1+y)=(7-y)(1+y)#

#rArr16-(y-3)^2=(7-y)(1+y)#