{"id":10806,"date":"2022-05-30T15:40:45","date_gmt":"2022-05-30T10:10:45","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10806"},"modified":"2022-05-30T15:40:48","modified_gmt":"2022-05-30T10:10:48","slug":"verify-that-the-numbers-given-alongside-of-the-cubic-polynomials-below-are-their-zeroes-also-verify-the-relationship-between-the-zeroes-and-the-coefficient-in-each-case","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/verify-that-the-numbers-given-alongside-of-the-cubic-polynomials-below-are-their-zeroes-also-verify-the-relationship-between-the-zeroes-and-the-coefficient-in-each-case\/","title":{"rendered":"Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficient in each case :"},"content":{"rendered":"
Question<\/strong> : Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficient in each case :<\/p>\n (i)<\/strong>\u00a0 \\(2x^3 + x^2 – 5x + 2\\);\u00a0 \\(1\\over 2\\), 1, -2<\/p>\n (ii)<\/strong>\u00a0 \\(x^3 – 4x^2 + 5x – 2\\);\u00a0 2, 1, 1<\/p>\n Solution<\/strong> :<\/p>\n (i)<\/strong>\u00a0 Comparing the given polynomials with \\(ax^3 + bx^2 + cx + d\\), we get<\/p>\n a = 2, b = 1, c = -5 and d = 2<\/p>\n p(\\(1\\over 2\\)) = 2\\(({1\\over 2})^3\\) + \\(({1\\over 2})^2\\) – 5(\\(1\\over 2\\)) + 2 = \\(1\\over 4\\) + \\(1\\over 4\\) – \\(5\\over 2\\) + 2 = \\(0\\over 4\\) = 0<\/p>\n p(1) = 2\\((1)^3\\) + \\((1)^2\\) – 5(1) + 2 = 2 + 1 – 5 + 2 = 0<\/p>\n p(-2) = 2\\((-2)^3\\) + \\((-2)^2 – 5(-2) + 2 = -16 + 16 = 0<\/p>\n \\(\\therefore\\)\u00a0 \u00a0\\(1\\over 2\\), 1 and -2<\/strong> are the zeroes of\u00a0 \\(2x^3 + x^2 – 5x + 2\\).<\/p>\n So,\u00a0 \\(\\alpha\\) = \\(1\\over 2\\), \\(\\beta\\) = 1 and \\(\\gamma\\) = -2.<\/p>\n Therefore, \\(\\alpha\\) + \\(\\beta\\) + \\(\\gamma\\) = \\(1\\over 2\\) + 1 + (-2) = \\(-1\\over 2\\) = \\(-b\\over a\\)<\/strong><\/p>\n \\(\\alpha\\)\\(\\beta\\) + \\(\\beta\\)\\(\\gamma\\) + \\(\\gamma\\)\\(\\alpha\\)\u00a0 = (\\(1\\over 2\\))(1) + (1)(-2) + (-2)(\\(1\\over 2\\)) = \\(-5\\over 2\\) = \\(c\\over a\\)<\/strong><\/p>\n and\u00a0 \\(\\alpha\\)\\(\\beta\\)\\(\\gamma\\) = \\(1\\over 2\\) \\(\\times\\) 1 \\(\\times\\) (-2) = \\(-2\\over 2\\) = \\(-d\\over a\\)<\/strong><\/p>\n (ii)<\/strong>\u00a0 Comparing the given polynomials with \\(ax^3 + bx^2 + cx + d\\), we get<\/p>\n a = 1, b = -4, c = 5 and d = -2<\/p>\n p(2) = \\((2)^3\\) – 4\\((2)^2\\) + 5(2) – 2 = 8 – 16 + 10 – 2 = 0<\/p>\n p(1) = \\((1)^3\\) – 4\\((1)^2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0<\/p>\n \\(\\therefore\\)\u00a0 \u00a02, 1 and 1<\/strong> are the zeroes of \\(x^3 – 4x^2 + 5x – 2\\)<\/p>\n So,\u00a0 \\(\\alpha\\) = 2, \\(\\beta\\) = 1 and \\(\\gamma\\) = 1.<\/p>\n Therefore, \\(\\alpha\\) + \\(\\beta\\) + \\(\\gamma\\) = 2 + 1 + 1 = \\(-(-4)\\over 1\\) = \\(-b\\over a\\)<\/strong><\/p>\n \\(\\alpha\\)\\(\\beta\\) + \\(\\beta\\)\\(\\gamma\\) + \\(\\gamma\\)\\(\\alpha\\)\u00a0 = (2)(1) + (1)(1) + (1)(2) = \\(5\\over 1\\) = \\(c\\over a\\)<\/strong><\/p>\n and\u00a0 \\(\\alpha\\)\\(\\beta\\)\\(\\gamma\\) = 2 \\(\\times\\) 1 \\(\\times\\) 1 = -(-2) = \\(-d\\over a\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Question : Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficient in each case : (i)\u00a0 \\(2x^3 + x^2 – 5x + 2\\);\u00a0 \\(1\\over 2\\), 1, -2 (ii)\u00a0 \\(x^3 – 4x^2 + 5x – 2\\);\u00a0 2, 1, 1 Solution : …<\/p>\n