If the product of the zeroes of the polynomial (ax²-6x-6) is 4, then find the value of a.

Write whether the following expressions are polynomials or nt. Give reasons for your answer.
(i) \(x^{ 3 }+\frac { 1 }{ x^{ 2 } } +\frac { 1 }{ x } +1\)
(ii) x²+x+3
(iii) y^-12-3y+2
(iv) \(\sqrt { 2 } y^{ 3 }+\sqrt { 3 } y\)

If one zero of the polynomial (a²+9)x²+13x+6a is a reciprocal of the other, then find the value of a.

For a quadratic polynomial, whose one zero is 8 and the product of zeroes is -56.

Find all the zeroes of f(x) = x² - 2x

Find the quadratic polynomial whose sum and product of the zeroes are \(\frac{21}{8}\) and \(\frac{5}{16}\) respectively.

Find the condition that zeroes of polynomial p(x) = ax² + bx + c are reciprocal of each other.

If \(\alpha\) and \(\beta\) are the zeroes of a polynomial \({ x }^{ 2 }-4\sqrt { 3 } x+3\), then find the value of \(\alpha+\beta-\alpha\beta\)

Write the degree of the following polynomials.
\(7q^{ 6 }+4q^{ 2 }+\frac { 3 }{ 2 } +q-8\)

If 2 is a zero of polynomial p(x)=4x²+2x-5a, then find the value of a.

If a and β are the zeroes of the quadratic polynomial f(x)=ax^2+bx+c, then find the difference between the zeroes.
\(\left( \alpha -\beta \right) ^{ 2 }=\left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta \quad or\quad \left( \alpha -\beta \right) =\pm \sqrt { \left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta } \)

Find the zeroes of the given polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials \(7y^{ 2 }-\frac { 11 }{ 3 } Y-\frac { 2 }{ 3 } \)

Find all the zeroes of 2x⁴-3x³-3x²+6x-2, if you know that two of its zeroes are \(\sqrt { 2 } \) and \(-\sqrt { 2 } \) .

In the given figure, the graph of a polynomial p(x) is shown. Find the number of zeros of P(x).

If zeroes of the polynomial x² + 4x + 2a are \(\alpha\) and \(\frac{2}{\alpha}\), then find the value of a.

If \(\alpha\) and \(\beta\) are the zeroes of the polynomial x² + 8x + 6 from a quadratic polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)