{"id":10791,"date":"2022-05-30T01:41:32","date_gmt":"2022-05-29T20:11:32","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10791"},"modified":"2022-05-30T01:56:54","modified_gmt":"2022-05-29T20:26:54","slug":"obtain-all-the-zeroes-of-3x4-6x3-2x2-10x-5-if-two-of-its-zeroes-are-sqrt5over-3-and-sqrt5over-3","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/obtain-all-the-zeroes-of-3x4-6x3-2x2-10x-5-if-two-of-its-zeroes-are-sqrt5over-3-and-sqrt5over-3\/","title":{"rendered":"Obtain all the zeroes of \\(3x^4 + 6x^3 – 2x^2 – 10x – 5\\), if two of its zeroes are \\(\\sqrt{5\\over 3}\\) and -\\(\\sqrt{5\\over 3}\\)."},"content":{"rendered":"
Since two zeroes are \\(\\sqrt{5\\over 3}\\) and -\\(\\sqrt{5\\over 3}\\),<\/p>\n
x = \\(\\sqrt{5\\over 3}\\) and x = -\\(\\sqrt{5\\over 3}\\)<\/p>\n
\\(\\implies\\) (x – \\(\\sqrt{5\\over 3}\\))(x + \\(\\sqrt{5\\over 3}\\)) = \\(3x^2 – 5\\) is a factor of the given polynomial. Now, we apply the division algorithm to the given polynomial and \\(3x^2 – 5\\).<\/p>\n
<\/p>\n
First term of quotient is \\(3x^4\\over 3x^2\\) = \\(x^2\\)<\/p>\n
Second term of quotient is \\(6x^3\\over 3x^2\\) = 2x<\/p>\n
Third term of the quotient is \\(3x^2\\over 3x^2\\) = 1<\/p>\n
So, \\(3x^4 + 6x^3 – 2x^2 – 10x – 5\\) = (\\(3x^2 – 5\\))(\\(x^2 + 2x + 1\\)) + 0 = (\\(3x^2 – 5\\))\\({(x + 1)}^2\\)<\/p>\n
Quotient = \\(x^2 + 2x + 1\\) = \\({(x + 1)}^2\\)<\/p>\n
Zeroes of \\({(x + 1)}^2\\) are -1 and -1.<\/p>\n
Hence, all its zeroes are \\(\\sqrt{5\\over 3}\\), -\\(\\sqrt{5\\over 3}\\), -1, -1.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Since two zeroes are \\(\\sqrt{5\\over 3}\\) and -\\(\\sqrt{5\\over 3}\\), x = \\(\\sqrt{5\\over 3}\\) and x = -\\(\\sqrt{5\\over 3}\\) \\(\\implies\\) (x – \\(\\sqrt{5\\over 3}\\))(x + \\(\\sqrt{5\\over 3}\\)) = \\(3x^2 – 5\\) is a factor of the given polynomial. Now, we apply the division algorithm to the given polynomial and \\(3x^2 – 5\\). First term …<\/p>\n