Question

If two zeros of the polynomial f(x) =x^3-4x^2-3x+12 are โ3 and-โ3, then find its third zero give your answer correctly otherwise the answer will be reported

Answer 1

$(x - \sqrt{3} )(x + \sqrt{3} )$
are the two factors of polynomial.on Multiplication of these two
${x}^{2} - 3$
now divide the polynomial by this factor ,and from quotient we will find another zero
${x}^{2} - 3) {x}^{3} - 4 {x}^{2} - 3x + 12(x - 4 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} \: \: \: \: \: \: \: \: \: \: \: \: - 3x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 4 {x}^{2} \: \: \: \: \: \: \: \: \: \: + 12 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 4 {x}^{2} \: \: \: \: \: \: \: \: \: \: \: + 12 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \\ - - - - - - - - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0$
the quotient is
$x - 4$
so to find zero
$x - 4 = 0 \\ x = 4$
is the third zero.
please do request for answer.don't give order.