How do you factor #3x^2 +10x - 8#?

2 Answers
Apr 1, 2018

#(3x-2)(x+4)#

Explanation:

#3x^2+10x-8#

first multiply #3# with #-8# and we get #-24#

then find two numbers that if we add or subtract we get #+10# and if we multiply we get #-24#

and the numbers are #+12# and #-2#

then we place those numbers as follow :-

#3x^2+12x-2x-8#

then we group them

#(3x^2+12x)(-2x-8)#

now take the common factor out.

#3x(x+4)-2(x+4)#

then group the numbers outside of the brackets

#(3x-2)(x+4)#

and that is the answer

Proper Solution:-

#3x^2+10x-8#

#3x^2+12x-2x-8#

#(3x^2+12x)(-2x-8)#

#3x(x+4)-2(x+4)#

#(3x-2)(x+4)#

Apr 1, 2018

#(x+4)(3x-2)#

Explanation:

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

#•color(white)(x)ax^2+bx+c ;a!=0#

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

#3x^2+10x-8larrcolor(blue)"is in standard form"#

#"with "a=3,b=10" and "c=-8#

#"consider factors of "3xx-8=-24" which sum to + 10"#

#"the required factors are "+12" and "-2#

#color(blue)"split the middle term using these factors"#

#3x^2+12x-2x-8larrcolor(blue)"factor by grouping"#

#=color(red)(3x)(x+4)color(red)(-2)(x+4)#

#"take out the common factor "(x+4)#

#=(x+4)(color(red)(3x-2))#

#rArr3x^2+10x-8=(x+4)(3x-2)#