Question

Show that
$\frac{1}{3}$
and
$\frac{1}{4}$
are the zeroes of the polynomial
$9 {x}^{3} - 6 {x}^{2} - 11x + 4$
Also find the 3rd zero of the polynomial.

Please help with answer...
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Answer 1

Answer:

put x= 1/3

9(1/27) -6( 1/9) -11/3 +4,

1/3 - 2/3 -11/3 +4

-12/3 +4

-4 +4 = 0

so 1/3 is zero

similarly 9(1/64) -6/16 -11/4 +4

9/64 - 3/8 -11/4 +4

(9 -24 -176 +256)/64

=0

3rd zero

sum of zeroes = 6/9 = 2/3

so 3rd zero = 2/3 - 1/3 -1/4

= 1/12