What is the complete factorization of #x^3+8x^2+17x+10#?

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Answer 1 · 2017-06-29T06:55:24

#x^3+8x^2+17x+10=(x+1)(x+2)(x+5)#

Let #x^3+8x^2+17x+10=(x+a)(x+b)(x+c)#

then #abc=10# and hence zeros are factors of #10# i.e. #+-1,+-2,+-5,+-10#

As we have all positive signs #1,2,5,10# are not the zeros of the polynomial. Let us try #-1# and #-2# and both make the polynomial zero.

as #(-1)^3+8(-1)^2+17(-1)+10=-1+8-17+10=0# and

#(-2)^3+8(-2)^2+17(-2)+10=-8+32-34+10=0#

Hence #(x+1)# and#(x+2)# are two factors of #x^3+8x^2+17x+10#

third factor can only be #(x+5)# (as #abc=10#) so let us check for #-5# too

#(-5)^3+8(-5)^2+17(-5)+10=-125+200-85+10=0#

Hence #x^3+8x^2+17x+10=(x+1)(x+2)(x+5)#