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Question:I have 2 equations, I need to know the Factors and crucially, how you got them, preferably using algebraic division to answer, the first:x^3-3x^2+4 I know that one factor is (x+1) I just need the other two the Other: x^3-3x^2-x-4 any help will be much appreciated, Thanks!! preferably in the form (x-a)(x-b)(x-c) and algebreic division to form a quadratic answer, which can be expanded to this.

Answers:1) = x^3-3x^2+4 = (x^3+x^2) + ( -4x^2-4x) + (4x+4) = (x+1)(x^2+x) -4x(x+1) + 4 (x+1) = (x+1)(x^2+x-4x+4) = (x+1)(x^2 -3x +4) There are no other factors since for 2nd equation (x^2 -3x +4), (D) = b^2 - 4ac = 9-16 is less than zero 2) For second question i have used an online utility at http://id.mind.net/~zona/ezGraph/ezGraph.html So it is clear that there is only one factor for this equation too which lies between 3 & 4. It is very hard to solve a polynomial of this kind by hand. In general a program will work fine for you. Approx solution of this is x = 3.589

Question:The base of a pyramid covers an area of 13.0 acres (1acre = 43,560 square feet) and has a height of 481 feet. If the volume of a pyramid is given by the expression V = (1/3)(b)(h) where b is the area of the base and h is the height is the height, find the area in cubic meters. That's what the problem from my book. I tried the problem and got 2.5708 x 10^6 meters cubed, but I'm not sure if I'm right. Can anyone Tell me if I have one too many significant numbers in my answer or tell me if my answer is just completely wrong? Thanks in advance.

Answers:Answer --> 2.57E+6 m^3 Given in SI units: b = 52,608.73 m^3 h = 146.6088 m Using your equation: V = (1/3) * b * h V = (1/3) * (52,608.73 m^3) * (146.6088 m) V = 2,570,968 m^3 So 2.57E+6 m^3

Question:I need to find equation of a cubic (ax^3+bx^2+cx+d) with the local max (-2,5) and inflection point (0,1) given. A little help? Thanks.

Answers:Let be f(x) = ax^3+bx^2+cx+d If for x = 0 it's inflection point then x = 0 is zeros of the second derivative of f(x) the second derivative f ' ' (x) = 6ax +2b f '' (0) = 2b then b = 0 and f (0) = 1 then d = 1 Now write f (x) = ax^3 + cx+1 if f(x) has a local max (-2,5) then x = -2 it's a zero of first derivative f ' (x) = 3ax^2+c f ' (-2 ) = 12a+c = 0 now c = -12a and f (- 2) = -8a-2c+1 = 5 then -8a +24a = 4 a = 1/4 c = - 3 then f(x) = (1/4)x^3 - 3x+1

Question:A cubic equation may be expressed as ax^3 + bx^2 +cx +d = 0 or as (x - w)(x - y)(x - z) = 0 where x,y,z are roots of the equation. Use this fact to find the values of (w+y+z), (wy+wz+yz) and (wyz) in terms of a,b,c and d I have no idea where to start. Please help me!!! thanks in advance, Steve

Answers:wyz = -d //The constant term in the cubic is always the product of the constants in the factored form. wy + wz + yz = -c //This is sum of roots taken two at a time which is the negative coefficient of the x term in the cubic. w+y+z = -b //The sum of the roots is the negative coefficient of the x^2 term. Hope this helped.

From Youtube

cubic / quartic equations II :The process for solving depressed cubics by reducing it to a system of equations which becomes an 'easy' sixth degree equation.

Solving Cubic Equations :www.gdawgenterprises.com This video demonstrates how to solve cubic equations, or in other words, how to solve equations with third degree polynomials. The methods shown are the cube root method, graphing, and factoring using long division. It is shown how a cubic expression may consist of the product of three binomials.