How do you factor and simplify #4sin^2y+8siny+4#?

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Answer 1 · 2016-09-04T16:45:50

#4 (sin(y) + 1)^(2)#

We have: #4 sin^(2)(y) + 8 sin(y) + 4#

This expression can be factorised using the middle-term break:

#= 4 sin^(2)(y) + 4 sin(y) + 4 sin(y) + 4#

#= 4 sin(y) (sin(y) + 1) + 4 (sin(y) + 1)#

#= (sin(y) + 1) (4 sin(y) + 4)#

We can simplify this expression further by factoring out a #4# in the second set of parentheses:

#= (sin(y) + 1) (4 (sin(y) + 1))#

#= 4 (sin(y) + 1) (sin(y) + 1)#

#= 4 (sin(y) + 1)^(2)#