How do you factor #2x^2-11x-21#?

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Answer 1 · 2015-05-17T16:32:00

By finding its roots and, then, factoring your function!

Using Bhaskara to find its roots:

#(11+-sqrt(11^2-4(2)(-21)))/(2*2)#
#(11+-17)/4#
#x=7#, which is the same as the factor #x-7=0#
and #x=-3/2#, which is the same as the factor #2x+3=0#

Now, you just need to reaggregate them:

#(x-7)(2x+3)#

Answer 2 · 2015-05-17T16:47:41

#2 x^2 - 11 x - 21# can be factored as #(x - 7)(2x + 3)#, as follows:

#2x^2 - 11 x - 21# #=# #2x^2 - 14 x + 3 x - 21# #=# #(2 x^2 - 14 x) + (3 x - 21)# #=# #2 x (x - 7) + 3(x - 7)# #=# #(x - 7)(2x + 3)#