Question

√3 and -√3 are zeroes of f(x) = x4-3x³-x²+9x-6 find all the zeroes of p(x)
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Answer 1

$f( \sqrt{3}) = { \sqrt{3} }^{4} - 3 \times { \sqrt{3} }^{3} - { \sqrt{3} }^{2} + 9 \times \sqrt{3} - 6$

$9 - 9 \sqrt{3} - 3 + 9 \sqrt{3} - 6 \: = 0$

$f( \sqrt{3} ) = { - \sqrt{3} }^{4} - 3 \times { - \sqrt{3} }^{3} - { - \sqrt{3} }^{2} + 9 \times - \sqrt{3} - 6$

$9 + 9 \sqrt{3 } - 3 - 9 \sqrt{3} - 6 = 0$

Answer 2

Answer:

Step-by-step explanation: