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(x^2 + 2y)(x^3 - 2xy + y^3) = x^2(x^3 - 2xy + y^3) + 2y(x^3 - 2xy + y^3)
Now, let's distribute each term:
= x^2 * x^3 - x^2 * 2xy + x^2 * y^3 + 2y * x^3 - 2y * 2xy + 2y * y^3
Now, simplify each term:
= x^5 - 2x^3y + x^2y^3 + 2x^3y - 4xy^2 + 2y^4
Now, combine like terms:
= x^5 + x^2y^3 - 4xy^2 + 2y^4
So, the expanded form of (x^2 + 2y)(x^3 - 2xy + y^3) is x^5 + x^2y^3 - 4xy^2 + 2y^4.
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Can anyone help you factor 81ax3-162ax2y plus 108axy2-24ay3?
3a(3x - 2y)(3x - 2y)(3x - 2y) or 3a times (3x - 2y) to the third power
What is the terms that include the same variable?
Not sure. There is no particular name for x2 and 2xy (both include x) in x2 + 2xy + y2 + y3
How do you factor x to the sixth minus y to the sixth?
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
How do you factor x3 plus y3?
(x2 - xy + y2)(x + y)
How to factirise this x3 y3 equation?
x3-y3
How do you write median median line?
the equation is A= y2-y2/x3-x2 after that you find the y-intercept by doing, b= y1+y2+y3-A(x1+x2+x3)/3
Given three points x1y1x2y2 x3y3 write a program to check if all three points fall on one straight line using only conditional statementsififelse nested if only?
#include<stdio.h> #include<conio.h> void main() { int x1, x2, x3, y1, y2, y3 ; printf("\nEnter the Co-ordinates of first point(x1,y1)"); scanf("d",&1x, &y1); printf("\nEnter the Co-ordinates of 2nd point(x2,y2)"); scanf("d",&x2, &y2); printf("\nEnter the Co-ordinates of 3rd point(x3,y3)"); scanf("d",&x3, &y3); if( (y2-y1)/(x2-x1)==(y3-y2)/(x3-x2) ) printf("The three points lie on straight line"); else printf("The three points do not lie on straight line"); getch(); }
What is the answer to x plus y to 3rd power plus x minus y to 3rd power?
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
How do you compute quadratic equation using java?
What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
What is the answer of x2 plus y3 plus 15?
X2+Y3+15 All you can do is simplify it.
How do you find the midpoint of a scalene triangle?
It makes no difference that the triangle is scalene. If the coordinates of the three vertices are (x1, y1), (x2, y2) and (x3, y3), then the coordinates of the midpoint (centroid) is [(x1 + x2 + x3)/3, (y1 + y2 + y3)/3]. Alternatively, join any two vertices to the midpoint of the opposite side. They (and the third median) will meet at the centroid.
When can a polynomial is said to be sum or difference of two cubes?
When it is of the form x3 + y3 or x3 - y3. x or y can have coefficients that are perfect cubes, or even ratios of perfect cubes eg x3 + (8/27)y3.
