How do you solve #5b ^ { 3} + 15b ^ { 2} + 12b= - 36#?

1 Answer
Jim G.
Oct 25, 2017

#b=-3,b=+-(2sqrt15)/5i#

Explanation:

#"rearrange equating to zero"#

#rArr5b^3+15b^2+12b+36=0#

#"factorise by 'grouping'"#

#=color(red)(5b^2)(b+3)color(red)(+12)(b+3)=0#

#"factor out "(b+3)#

#rArr(b+3)(color(red)(5b^2+12))=0#

#"equate each factor to zero and solve for b"#

#b+3=0rArrb=-3larrcolor(blue)"real root"#

#5b^2+12=0rArrb^2=-12/5#

#rArrb=+-sqrt(-12/5)#

#color(white)(rArrb)=+-(2sqrt3)/(sqrt5)i#

#color(white)(rArrb)=+-(2sqrt15)/(5)ilarrcolor(blue)" complex roots"#

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