How do you factor the trinomial #2x^2+10x+25#?

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Answer 1 · 2017-12-17T01:34:41

#x=(-5+5i)/2,##(-5-5i)/2#

#2x^2+10x+25# is a quadratic equation in standard form:

#ax^2+bx+c#,

where:

#a=2#, #b=10#, and #c=25#

Set the equation to #0# and solve for #x#.

#0=2x^2+10x+25#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values.

#x=(-10+-sqrt(10^2-4*2*25))/(2*2)#

Simplify.

#x=(-10+-sqrt(-100))/4#

Prime factorize the square root.

#x=(-10+-sqrt(-2xx2xx5xx5))/4#

Simplify.

#x=(-10+-10i)/4#

Factor out the common #2# in the numerator.

#x=(2(-5+-5i))/4#.

Divide #4# in the denominator by the #2# in the numerator.

#x=(-5+-5i)/2#

Solutions for #x#.

#x=(-5+5i)/2,##(-5-5i)/2#