{"id":10814,"date":"2022-05-30T15:51:38","date_gmt":"2022-05-30T10:21:38","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10814"},"modified":"2022-05-30T15:51:41","modified_gmt":"2022-05-30T10:21:41","slug":"find-the-cubic-polynomial-with-the-sum-sum-of-the-products-of-its-zeroes-taken-two-at-a-time-and-the-product-of-its-zeroes-as-2-7-14-respectively","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-cubic-polynomial-with-the-sum-sum-of-the-products-of-its-zeroes-taken-two-at-a-time-and-the-product-of-its-zeroes-as-2-7-14-respectively\/","title":{"rendered":"Find the cubic polynomial with the sum, sum of the products of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, -14 respectively."},"content":{"rendered":"
Let the cubic polynomial be \\(ax^3 + bx^2 + cx + d\\), and its zeroes be \\(\\alpha\\), \\(\\beta\\) and \\(\\gamma\\).<\/p>\n
Then,\u00a0 \\(\\alpha\\) + \\(\\beta\\) + \\(\\gamma\\) = 2 = \\(-(-2)\\over 1\\) = \\(-b\\over a\\)<\/p>\n
\\(\\alpha\\)\\(\\beta\\) + \\(\\beta\\)\\(\\gamma\\) +\u00a0\\(\\gamma\\)\\(\\alpha\\) = – 7 = \\(-7\\over 1\\) = \\(c\\over a\\)<\/p>\n
and \\(\\alpha\\)\\(\\beta\\)\\(\\gamma\\) = -14 = \\(-14\\over 1\\) = \\(-d\\over a\\)<\/p>\n
If a = 1, then b = -2, c = -7 and d = 14.<\/p>\n
So, one cubic polynomial which satisfy the given conditions will be \\(x^3 – 7x + 14\\)<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Let the cubic polynomial be \\(ax^3 + bx^2 + cx + d\\), and its zeroes be \\(\\alpha\\), \\(\\beta\\) and \\(\\gamma\\). Then,\u00a0 \\(\\alpha\\) + \\(\\beta\\) + \\(\\gamma\\) = 2 = \\(-(-2)\\over 1\\) = \\(-b\\over a\\) \\(\\alpha\\)\\(\\beta\\) + \\(\\beta\\)\\(\\gamma\\) +\u00a0\\(\\gamma\\)\\(\\alpha\\) = – 7 = \\(-7\\over 1\\) = \\(c\\over a\\) and \\(\\alpha\\)\\(\\beta\\)\\(\\gamma\\) = -14 = \\(-14\\over …<\/p>\n