What are all the possible rational zeros for #f(x)=x^3+13x^2+38x+24# and how do you find all zeros?

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Answer 1 · 2017-09-16T12:33:11

Zeros are #-3 , -9.123, -0.877#

Possible rational zeros are found as under ,

Factors of constant term #24# are #+-( 1,2,3,4,6,8,12,24)/#

factor of leading coefficient #1# is #+-1# , so possible

rational zeros are #+-( 1,2,3,4,6,8,12,24)#. By trying long divisions

# -3 # suits only . #f(x) = x^3 +13x^2 +38x+24 # or

#f(x) = x^3 +3x^2 +10x^2+30x+8x+24 # or

#f(x) = x^2(x+3)+ 10x(x+3)+8(x+3) # or

#f(x) = (x+3)(x^2+10x +8)#, for # (x^2+10x +8)#

#x= (-10 +- sqrt( 100 -32))/2 # or

#x ~~ -9.123 , x ~~ -0.877 (3dp) :.#

#f(x) = x^3 +13x^2 +38x+24 = (x+3)(x+9.123)(x+0.877)#

Zeros are #-3 , -9.123, -0.877# [Ans]