Given #z(x) = 12x^3 + kx^2 - 131x-105#, #z(-2) = 77#, and #z(-3) = 0#, what are all the zeros of #z(x)#?

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Answer 1 · 2016-08-16T13:18:50

All the zeroes of #z(x)# are # -3, -5/6, and, 7/2#.

#z(x)=12x^3+kx^2-131x-105, and, z(-3)=0#

#rArr 12(-27)+k(9)-131(-3)-105=0#, i.e.,

#-324+9k+393-105=0, or, 9k=36 rArr k=4#.

Therefore, #z(x)=12x^3+4x^2-131x-105#.

We are given that #z(-3)=0 rArr (x+3)# is a factor of #z(x)#.

Now, #z(x)=12x^3+4x^2-131x-105#

#=ul(12x^3+36x^2)-ul(32x^2-96x)-ul(35x-105)#

#=12x^2(x+3)-32x(x+3)-35(x+3)#

#=(x+3)(12x^2-32x-35)#

#=(x+3){ul(12x^2-42x)+ul(10x-35)}#

#=(x+3){6x(2x-7)+5(2x-7)}#

#=(x+3)(2x-7)(6x+5)#

Since, #z(x)# is a cubic poly., it can not have more than #3# zeroes.

Therefore, all the zeroes of #x(x)# are # -3, -5/6, and, 7/2#.

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