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How do you factor this equation using the ac method 4x3-19x2 plus 15x?
Wiki User
∙ 8y ago
x(x - 1)(4x - 15)
Wiki User
∙ 8y ago
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What are the solutions of the equation x3 plus 3x2-x-3 equals 0?
the solutions to this equation are -1,+1 and -3. you can solve this equation by using the polynomial long division method. we basically want to factorize this and polynomial and equate its factors to zero and obtain the roots of the equation. By hit and trial , it clear that x=1 i.e is a root of this equation. So (x-1) should be a factor of the given polynomial (LHS). Divide the polynomial by x-1 using long division method and you will get the quotient as x2+4x+3 and remainder would be 0 ( it should be 0 as we are dividing the polynomial with its factor. Eg when 8 is divided by any of its factor like 4,2 .. remainder is always zero ) Now, we can write the given polynomial as product of its factors as x3+3x2-x-3 = (x-1)(x2+4x+3) =(x-1)(x+1)(x+3) [by splitting middle term method] so the solutions for the given polynomial are obtained when RHS = 0, Hence x=-1 , X = +1, x=-3 are the solutions for this equation.
When you factor using the Zero Product Rule the solutions to the simpler equations are also the solutions to the original equation?
True yal :)
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Coness
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using the quadratic formula or the graphics calculator. Yes, you can do it another way, by using a new method, called Diagonal Sum Method, that can quickly and directly give the 2 roots, without having to factor the equation. This method is fast, convenient and is applicable to any quadratic equation in standard form ax^2 +bx + c = 0, whenever it can be factored. It requires fewer permutations than the factoring method does, especially when the constants a, b, and c are large numbers. If this method fails to get answer, then consequently, the quadratic formula must be used to solve the given equation. It is a trial-and-error method, same as the factoring method, that usually takes fewer than 3 trials to solve any quadratic equation. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009)
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ng ffh yc bub ajgf bcjsgfk ifydbtdesb unhhb
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Start with a quadratic equation in the form � � 2 � � � = 0 ax 2 +bx+c=0, where � a, � b, and � c are constants, and � a is not equal to zero ( � ≠ 0 a =0).
What are the solutions of the equation x3 plus 3x2-x-3 equals 0?
the solutions to this equation are -1,+1 and -3. you can solve this equation by using the polynomial long division method. we basically want to factorize this and polynomial and equate its factors to zero and obtain the roots of the equation. By hit and trial , it clear that x=1 i.e is a root of this equation. So (x-1) should be a factor of the given polynomial (LHS). Divide the polynomial by x-1 using long division method and you will get the quotient as x2+4x+3 and remainder would be 0 ( it should be 0 as we are dividing the polynomial with its factor. Eg when 8 is divided by any of its factor like 4,2 .. remainder is always zero ) Now, we can write the given polynomial as product of its factors as x3+3x2-x-3 = (x-1)(x2+4x+3) =(x-1)(x+1)(x+3) [by splitting middle term method] so the solutions for the given polynomial are obtained when RHS = 0, Hence x=-1 , X = +1, x=-3 are the solutions for this equation.
