How do you completely factor #25x^3+150x^2+131x+30# given 5x+3 is one of the factors?

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Answer 1 · 2016-10-04T19:10:57

#(5x+3)(5x+2)(x+5)#.

Let, #p(x)=25x^3+150x^2+131x+30.#

We are given that, #(5x+3)# is a factor of #p(x)#.

To factorise #p(x)# completely, Synthetic Division is usually

performed, but, we can proceed as under :-

#p(x)=25x^3+150x^2+131x+30#

#=ul(25x^3+15x^2)+ul(135x^2+81x)+ul(50x+30)#

#=5x^2(5x+3)+27x(5x+3)+10(5x+3)#

#=(5x+3)(5x^2+27x+10)#

#=(5x+3){ul(5x^2+25x)+ul(2x+10)}#
....................................................................#[25xx2=50, 25+2=27]#

#=(5x+3){5x(x+5)+2(x+5)}#

#:. p(x)=(5x+3)(5x+2)(x+5)#.