How do you factor #10x^2+29x+10#?

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Answer 1 · 2017-02-11T13:11:35

#(2x+5)(5x+2)#

First let's find the zeros of the polynomial by the quadratic formula:

#x=(-29+-sqrt(29^2-4*10*10))/(2*10)#

#=(-29+-sqrt441)/20#

#=(-29+-21)/20#

#x_1=(-29-21)/20=-5/2#

#x_2=(-29+21)/20=-2/5#

Since

#ax^2+by+c=a(x-x_1)(x-x_2)#,

you get

#10x^2+29x+10=10(x+5/2)(x+2/5)=2*(x+5/2)* 5*(x+2/5)=(2x+5)(5x+2)#