How do you factor #16z^{2} + 57= 64z#?

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Answer 1 · 2017-06-08T04:46:36

#(x-2-sqrt(7)/2)(x-2+sqrt(7)/2)=0#

The way you have written the question, there are two ways to factor this equation. First is to factor out #16z#.

#16z^2-64z+57=0#

#16z(z-4)+57=0#

The second is to use the quadratic equation and get the factorization with square roots in it.

#16z^2-64z+57=0#

#x=(-(-64)+-sqrt((-64)^2-4(16)(57)))/(2(16))#

#=(64+-sqrt(4096-3648))/(32)#

#=(64+-sqrt(448))/32#

#=(64+-sqrt(64xx7))/32#

#=(64+-8sqrt(7))/32#

#=(8+-sqrt(7))/4#

#=2+-sqrt(7)/2#

These give the complete factors

#(x-2-sqrt(7)/2)(x-2+sqrt(7)/2)=0#