How do you factor the binomial #4x^2 + 15x + 10#?

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Answer 1 · 2015-06-14T19:02:54

Use quadratic formula to find zeros of #4x^2+15x+10# and hence factors:

#4x^2+15x+10 = (2x+(15+sqrt(65))/4)(2x+(15-sqrt(65))/4)#

#f(x) = 4x^2+15x+10# is of the form #ax^2+bx+c# with #a=4#, #b=15# and #c=10#

The discriminant is given by the formula:

#Delta = b^2-4ac = 15^2-(4xx4xx10) = 225 - 160 = 65#

This is positive but not a perfect square, so the roots of #f(x) = 0# and coefficients of the linear factors are real but irrational.

#f(x)=0# has roots given by the quadratic formula

#x = (-b +- sqrt(Delta))/(2a) = (-15+-sqrt(65))/8#

Since the leading term we are aiming for is #4x^2#, multiply this through by #2# to get:

#2x = (-15+-sqrt(65))/4#

Hence #f(x)# may be factored as:

#(2x+(15+sqrt(65))/4)(2x+(15-sqrt(65))/4)#