How do you solve #(8z+4)(5z+10)=0#?

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Answer 1 · 2016-12-21T23:13:36

To solve this, you can set each set of parenthesis equal to zero.

#8z+4=0# and #5z+10=0#

Solve for #z#

#8z+4=0#

Subtract #4# on both sides of the equation

#8z=-4#

Divide #8# on both sides of the equation

#z=-1/2#

Solve for #z#

#5z+10=0#

Subtract #10# on both sides of the equation

#5z=-10#

Divide both sides of the equation by #5#

#z=-2#

So, the solutions are:

#z=-1/2 and z=-2#

Answer 2 · 2016-12-21T23:22:24

#z=-1/2,z=-2#

#(8z+4)(5z+10)=0#
Zero Product Principle: If #ab=0# then #a=0#, #b=0# OR both equal zero.

Therefore, in terms of a factored polynomial, you would have to assume both have the chance of being equal to 0.

#(8z+4)=0# and #(5z+10)=0#...

#8z+4=0#
#8z\stackrel{\cancel{-4}}{\cancel{+4}}=0-4#
#8z=-4#
#z=-4/8=-1/2# or #-0.5#

#5z+10=0#
#5z\stackrel{\cancel{-10}}{\cancel{10}}=0-10#
#5z=-10#
#z=-10/5=-2#

answer
#z=-1/2,z=-2#