How do you solve #(5x - 25) ( x ^ { 2} + 10x + 21) = 0#?

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Answer 1 · 2017-04-21T22:33:10

See below.

Since this is already "factored," all we do is set each expression inside the parenthesis equal to zero.

#5x-25=0#
#5x=25#
#x=5#

#x^2+10x+21=0#

#x^2+7x+3x+21=0#

#x(x+7)+3(x+7)=0#

#(x+3)(x+7)=0#

So

#x=-3,-7#

Thus, our solutions for #x# are #-3,5,-7#.