tag:blogger.com,1999:blog-2433841880619171855.post5760782355926652828..comments2024-03-27T21:09:44.320+00:00Comments on Pat'sBlog: Synthetic Divison by a QuadraticUnknownnoreply@blogger.comBlogger10125tag:blogger.com,1999:blog-2433841880619171855.post-37922081567069198452022-10-03T14:34:25.108+01:002022-10-03T14:34:25.108+01:00excellent! the best...
excellent! the best...<br />Anonymoushttps://www.blogger.com/profile/00812856153188042635noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-16639896051217193342020-04-16T11:57:27.354+01:002020-04-16T11:57:27.354+01:00Wow! AmazingWow! AmazingErrorFinderhttps://www.blogger.com/profile/16174200817796249315noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-72092645674338216352019-09-05T04:18:48.859+01:002019-09-05T04:18:48.859+01:00how you can extend the use of synthetic division i...how you can extend the use of synthetic division in dividing polynomial by a quadratic expression of x² + bx + c?Anonymoushttps://www.blogger.com/profile/11756839586214307160noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-4731633648003020632019-08-23T04:05:02.872+01:002019-08-23T04:05:02.872+01:00What if the divisor was something like 3x^2+x+1 or...What if the divisor was something like 3x^2+x+1 or x^2-6x+9Anonymoushttps://www.blogger.com/profile/02002549515784862836noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-66157492042787326112019-07-06T17:09:57.540+01:002019-07-06T17:09:57.540+01:00Dear pat, this is daksh who wrote this comment as ...Dear pat, this is daksh who wrote this comment as "unknown". You can contact me here - dakshsinha7631@gmail.comAnonymoushttps://www.blogger.com/profile/12104824942256164999noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-6713070816964051072019-07-06T17:07:53.344+01:002019-07-06T17:07:53.344+01:00Thanks pat, it was really helpful I'll do a li...Thanks pat, it was really helpful I'll do a little more practice on it to find out more about it hope I'll get some help from you , if you have done some work on cubic one please send me so that we could do it together and may be we succeedAnonymoushttps://www.blogger.com/profile/12104824942256164999noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-43797950527549241932011-08-30T15:33:11.866+01:002011-08-30T15:33:11.866+01:00thank you so much! this blog was really a great he...thank you so much! this blog was really a great help...keep it up!Wakwakhttps://www.blogger.com/profile/04977938080524796839noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-59565528402496887632009-02-16T16:52:00.000+00:002009-02-16T16:52:00.000+00:00Sarah, I assume you can factor out a, and then div...Sarah, <BR/><BR/>I assume you can factor out <I>a</I>, and then divide the quotient by it.<BR/><BR/>Instead of dividing by 3x^2 - 5x + 6,<BR/>divide by x^2 - (5/3)x + 2, and then divide the quotient and remainder by 3.<BR/><BR/>In general I prefer long division to synthetic, but occasionally I'll use synthetic. In these cases though, isn't the familiarity with the algorithm from arithmetic a tremendous incentive to stick there?<BR/><BR/>JonathanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-35304419936340395232009-02-03T10:47:00.000+00:002009-02-03T10:47:00.000+00:00What do you do if one or both of the polynomials h...What do you do if one or both of the polynomials have a first number other than one.. like dividing by 2x^2 +x+1 or somthing.. what do I do with the 2?<BR/>Can those just not be done?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-50239451917617753662009-02-02T17:30:00.000+00:002009-02-02T17:30:00.000+00:00Mr. Ballew,would you explain what you mean by brea...Mr. Ballew,<BR/>would you explain what you mean by breaking up the cubic... is that another way to do it?Anonymousnoreply@blogger.com