Question

find the zeros of the polynomial 6x³ - 7x²- 11x + 12 , if x-1 is the factor of the polynomial

Answer 1

$since \: x - 1 \: is \: the \: one \: root \: of \: given \: equetion$
$therefore \: \frac{6x {}^{3} - 7x {}^{2} - 11x + 12 }{x - 1}$
$= 6x {}^{2} - x - 12 \:$
$6x {}^{2} - (9x + 8x) \: - 12$

$3x(2x - 3) + 4(2x - 3)$
$(3x + 4)(2x - 3)$
roots are
$(x - 1)(3x + 4)(2x - 3) = 0$
therefore x = 1, -4/3 , 3/2


hope it helps...

Answer 2

x-1=0

x=1

Putting the value in 6x^3-7x^2-11x+12

=6(1)^3-7(1)^2-11(1)+12

=6*1-7*1-11+12

=6-7-11+12

=18-18

=0