Find the zero of the polynomial 6x³-7x²-11x+12 if x-1 is a factor of the polynomial .
Answers
Answered by
4
Answer:
Step-by-step explanation:
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
Answered by
0
Answer:
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 and 3/2
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