 ! . U ? U ? U ? U ? – U ? in this class we will discuss cardon method for solution of a qb equation and discard as rule of sign. Cardon method of solution of qb equation. So, cardon method provides a technique for solving the general qb equation A x cube plus B x square plus C x plus D equal to 0. So, my cardon method or how I had am at a qb equation for modern prime example 1 solve the following equation x cube minus 15 x square minus 33 x plus 80 847 is equal to 0. Now given equation is x cube minus 15 x square minus 33 x plus 847 is equal to 0. So, here a 0 is equal to 1 a 1 is equal to minus 15 a 2 is equal to minus 33 and a 3 is equal to 147. So, first we have to remove the coefficient of x square to remove coefficient of x square we have to find is. So, h is equal to minus a 1 divided by n a 0. So, is equal to minus a 1 is equal to minus 15. So, divided minus 15 divided by 3 into 1. So, 15 by 3 is equal to 5. So, diminishing the roots of the equation by 5 by synthetic division we get now we apply synthetic division to the diminishing the roots of the equation by 5. So, 5 here coefficient of x cube is 1 coefficient of x square minus 15 coefficient of x minus 33 and constant term 847. So, here this one we take here then 5 into 1 5. So, minus 15 plus 5 minus 10. So, 5 into minus 10 minus 15 minus 33 plus minus 15 is equal to minus a 3 3. So, 5 into minus a 3 3 minus 415. Now, again here remainder will be 147 plus minus 1 minus 415 is equal to minus 433. Next this one we take here. So, 5 into 1 is minus 5 5 into 5 minus 10 plus 5 minus 5 5 into minus 5 minus 25 25. So, minus a 3 3 plus minus 25 is equal to minus 1 0 8. So, again this one we take here. So, 5 into 1 5. So, minus 5 plus 5 is equal to 0. So, here this term will be 0. So, x square term is removed. So, equation becomes the transform equation we get z cube minus 108 z plus 432 plus 432. So, this is the transform equation. Now, let z is equal to u plus e. So, cubing Gauchan we get z cube is equal to u plus v bracket all cube. So, u plus v all cube is equal to u cube plus v cube plus 3 u v bracket u plus v. So, here u plus v is equal to z. So, we put here z. So, u cube plus v cube plus 3 u v z. So, this term we take on 11 set. So, we get z cube minus 3 u v z minus bracket u cube plus v cube. So, now comparing equation 1 and 2 we get minus 3 u v equal to minus 108 and minus u cube plus v cube equal to 432. So, from here we get u v equal to 36 and u cube plus v cube is equal to minus 432. So, cubing this relation we get u cube plus v cube is equal to 36 to the power cube. Now, u cube and v cube are the roots of the equation e square plus 432 e plus 36 whole cube. So, from here we get t square plus. So, 432 we can write 2 into 6 cube t plus 36 whole cube we can write 6 cube whole square. So, from here we get t square plus 2 into 6 cube t plus 6 cube whole square. So, from here we get t plus 6 cube whole square. So, from here we get t plus 6 cube bracket t plus 6 cube bracket is equal to 0. So, here t is equal to minus 6 cube t is equal to minus 6 cube, but here u cube and v cube are the roots of this equation. So, we get u cube is equal to minus 6 cube v cube is equal to minus 6 cube. So, u cube is equal to minus 6 cube from here we get u is equal to minus 6 and v cube is equal to minus 6 cube we get v is equal to minus 6. So, here first here we have z is equal to u plus v. So, z is equal to minus 6 minus 6 is equal to minus 12. So, Rico Swarm is the root of the equation one. So we need to find other two roots so therefore I will buy synthetic division, So we applied synthetic division 11842 So we take one here So minus 12 into 1 we get minus 12, so 0 plus minus 12 minus 12. So minus 12 into minus 12 144, so we put 144 here. So minus 108 plus 144 we get 36, so minus 12 into 36 we get minus 43 2, so this will be 0, so remember 0. So we get, we factorize this equation of z q minus 1 0 is z plus 4 33 is going to z plus 12 break a z square minus 12 z plus 36. So, factorizing this equality equation we get z minus 6 into z minus 6. So, from here we get z is going to minus 12 6 6. So root of the z is equation root of the equation difference root of z are minus 12, 6, 6 and hence root of the equation is initially we take x is equal to z plus is therefore, we get z is equal to minus 12, 6, 6 so x will 5, that is minus 7, 11, 11. So, these are the roots of the equation. Now, we discuss this cartes rule of sign. The discard this rule of sign gives the information regarding the character and position of the roots of an equation. Let f of x is equal to 0 be an equation with real coefficient. So, first we find positive roots, an equation f of x with real coefficient cannot have more positive roots than the number of senses of signs from positive to negative or from negative to positive. Negative roots and equation f of x is equal to 0 cannot have more negative roots than the number of senses of sign in f of minus x is equal to 0. There are some important notes regarding this Cartes rule of sense. Number one, if the sign of the term of equation are all positive, then equation cannot have a positive root. Number two, if the equation is of even power of x and if all the coefficient have positive signs, it cannot have any real roots. If the equation of odd power of x and if all the term of positive and it has root 0 and no other real root. This Cartes rule of sense gives the maximum number of positive or negative roots of an equation, not the exact number. So, now we discuss three examples related to this Cartes rule of sense. Example two, find a lesser of the roots of the equation to x over 4 plus 3 x cube minus x square minus 1 is equal to 0. So, here number of senses is sign in f of x is plus plus minus plus minus. So, here senses of signs here plus positive to negative plus to minus. So, real sense is only one. So, equation f of x maximum one real roots of f of x is plus number of senses is sign in f of x is only one. Now, we find f of minus x. So, f of minus x will be 2 x square minus 3 x cube minus x square minus 1. So, here number of senses is sign in f of x here plus then minus minus minus. So, only one sense is there positive to negative. So, therefore, number of negative real roots of f of x is 1. So, here degree of this equation is 4. So, number of emerging roots is 4 minus 1 minus 1 is equal to 2. Next example three, find an upper limit of the real roots of the equation x over 4 minus 5 x cube plus 40 x square minus 8 x plus 24. So, it has 4 senses in sign 4 senses of sign plus minus plus minus plus. So, here is 4 senses plus to minus minus to plus plus to minus minus to plus. So, it has 4 senses of sign. So, equation has maximum number of positive real roots is 4. So, equation is of degree 4. So, upper limit of real roots is 4. Next example, example 4, apply this cartage rule of signs to a certain the minimum number of complete roots of the equation. Here given equation is x cube minus 3 x square minus 2 x minus 3. So, let f of x is equal to x cube minus 3 x square minus 2 x minus 3 is equal to 0. So, it has only one senses in sign. So, equation has maximum one positive real roots. The senses in sign f of x is only one. So, equation has maximum one positive root. Now, root. So, now, we find f of minus x. So, f of minus x will be x to the power 6 minus 3 x square plus 2 x minus 3 is equal to 0. So, here 3 senses in sign plus to minus minus to plus plus to minus. So, 3 senses in sign. So, given equation has utmost 3 negative roots. The given equation has utmost 1 plus 3 real roots, one positive 3 negative real roots. So, 4 real roots. So, minimum number of complex which is 6 minus 4. Here degree of equation is 6. So, 6 minus 4 is equal to 2. So, minimum number of complex which is 2. . . . . . . . .