How do you factor completely #6x^2+7x+2#?

2 Answers
Mar 28, 2018

Answer:

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

Explanation:

#"for a quadratic in "color(blue)"standard form"#

#•color(white)(x)ax^2+bx+c color(white)(x);a!=0#

#6x^2+7x+2larr"is in standard form"#

#"with "a=6, b=7" and "c=2#

#color(blue)"to factorise"#

#"consider the factors of the product ac which sum to b"#

#rArrac=6xx2=12" and + 4, + 3 sum to + 7"#

#6x^2+3x+4x+2larrcolor(blue)"split middle term"#

#=color(red)(3x)(2x+1)color(red)(+2)(2x+1)larrcolor(blue)"factor in groups"#

#"take out a common factor "(2x+1)#

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

#rArr6x^2+7x+2=(2x+1)(3x+2)#

Mar 28, 2018

Answer:

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

Explanation:

By sum & product

= #6x^2+4x+3x+2#

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

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

Hope this helps!