Polynomial Questions

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each given of the following :

Question : Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each given of the following : (i)  p(x) = \(x^3 – 3x^2 + 5x – 3\),  g(x) = \(x^2 – 2\) (ii)  p(x) = \(x^4 – 3x^2 + 4x + 5\), g(x) = \(x^2 + 1 – x\) …

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each given of the following : Read More »

Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.

Question : Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively. (i)  \(1\over 4\), -1 (ii)  \(\sqrt{2}\), \(1\over 3\) (iii)  0, \(\sqrt{5}\) (iv)  1, 1 (v)  \(-1\over 4\), \(1\over 4\) (vi)  4, 1 Solution : Let the polynomial be \(ax^2 + bx + c\) and its zeroes be \(\alpha\) and …

Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively. Read More »

Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its coefficients.

Question : Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its coefficients. (i)  \(x^2 – 2x – 8\) (ii)  \(4s^2 – 4s + 1\) (iii)  \(6x^2 – 3 – 7x\) (iv)  \(4u^2 + 8u\) (v)  \(t^2 – 15\) (vi)  \(3x^2 – x – 4\) Solution : (i)  \(x^2 – …

Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its coefficients. Read More »

The graph of y = p(x) are given below, for some polynomial p(x). Find the number of zeroes of p(x), in each case.

Question : The graph of y = p(x) are given below, for some polynomial p(x). Find the number of zeroes of p(x), in each case. Solution : (i)  There are no zeroes as the graph does not intersect the x-axis. (ii)  The number of zeroes is one as the graph intersect the x-axis at one point …

The graph of y = p(x) are given below, for some polynomial p(x). Find the number of zeroes of p(x), in each case. Read More »