How do you factor #42x^2+35x+7#?

2 Answers
Jun 17, 2018

#42*(x+1/3)(x+1/2)#

Explanation:

We solve the equation
#42x^2+35x+7=0#
dividing by #42#

#x^2+5/6x+1/6=0#
by the quadratic Formula we get

#x_(1,2)=-5/12pmsqrt(25/144-24/144)#
so we get

#x_1=-1/3# or #x_2=-1/2#

Jun 17, 2018

#7(2x+1)(3x+1)

Explanation:

#"take out a "color(blue)"common factor "7#

#=7(6x^2+5x+1)#

#"factor the quadratic using the a-c method"#

#"the factors of the product "6xx1=6#

#"which sum to + 5 are + 3 and + 2"#

#"split the middle term using these factors"#

#6x^2+3x+2x+1larrcolor(blue)"factor by grouping"#

#=color(red)(3x)(2x+1)color(red)(+1)(2x+1)#

#"take out the "color(blue)"common factor "(2x+1)#

#=(2x+1)(color(red)(3x+1))#

#42x^2+35x+7=7(2x+1)(3x+1)#