Please help me factor x^4 - y^4 - 2x^3y + 2xy^3.?
Group: (x^4 - y^4) - (2x^3y - 2xy^3) Use the rule for factoring the difference of squares on the left. Factor out the common factor on the right.: (x^2 + y^2) (x^2 - y^2) - 2xy (x^2 - y^2) Factor out the common factor (or use the distributive property, depending on how you look at it): (x^2 - y^2) (x^2 + y^2 - 2xy) Use the rule for factoring the difference of squares on the left. Rearrange the factor on the right: (x + y) (x - y) (x^2 - 2xy + y^2) Factor the right-most factor: (x + y) (x - y) (x - y)^2 Combine like terms: (x + y) (x - y)^3
How do you factor 2x^2+xy-y^2?
M: Think. The two factors must begin with 2X for one and just X for the other (2X=-?)(x +-?0; We note to have a negative Y^2 the second digit must be Y in each with one a + Y and the other a -Y. To have a +XY means 2X time Y must give us a +2Y in order for us to subtract a negative Y from it to get a single +XY. Voila. , The factors are (2X-Y)(X+Y)=2X^2 +2XY-XY-Y^2=2X^2 +XY-Y^2. Hope this helps. Good Luck
How do I factor this 15 {x} ^ {2} +2xy-24 {y} ^ {2}-x+24y-6?
Start by factoring [math]15x^2+2xy-24y^2[/math] and then hope for the best. Factor this by decomposition or by magic.Note that [math]15\times 24=360[/math]. So we seek two numbers with product being [math]360[/math] and difference [math]2[/math] with the larger assigned a positive sign. The numbers are [math]20[/math] and [math]18[/math]. So continue in the following way:[math]15x^2+2xy-24y^2=15x^2+20xy-18xy-24y^2=5x(3x+4y)-6y(3x+4y)=(5x-6y)((3x+4y).[/math]Now for the hoping part. The [math]-6[/math] could mess the whole thing up. In fact, had it been some other number, the quadratic would not factor. Here’s what we do. Set [math]y=0[/math] and factor the result. The factored form is [math](5x + 3)(3x - 2)[/math] with the important numbers being [math]3[/math] and [math]-2[/math]. This suggests that[math]15 {x} ^ {2} +2xy-24 {y} ^ {2}-x+24y-6=(5x-6y+3)(3x+4y-2)[/math].A quick check confirms.
How do I factorise x^2-2x-y^2+2y?
QUESTION:x^2–2x-y^2+2y1.Slightly arranging itx^2-y^2–2x+2y2.Adding yx and subtracting itx^2-y^2–2x+2y+yx-yx3.Rearranging the equationx^2+yx-2x-yx-y^2+2y4.Taking x and y, as they are commonx(x+y-2)-y(x+y-2)(x-y)(x+y-2)For clarification2.Adding yx and subtracting itIn this step yx is added, because the equation cannot be factorised as there are no common elements and hence yx is added so that it can be factorised.For eg.10= 10 +3–3Here 3 is added and then subtracted but 10 is stable and the total value of it is 10
Factor x^2+7x+y^2-7y-2xy-8 pls step by step?
x^2+7x+y^2-7y-2xy-8 = x^2 - 2xy + y^2 + 7 ( x - y ) - 8 ...............................= ( x - y )^2 + 7 ( x - y ) - 8 ...............................= ( x - y )^2 - ( x - y ) + 8 ( x - y ) - 8 ..............................= ( x - y ) [ x - y - 1 ] + 8 [ x - y - 1 ] ..............................= ( x - y + 8 ) ( x - y - 1 )
4x^4-17x^2y^2+4y^4 (Please factor!)?
→ (4x^2 - y^2)(x^2 - 4y^2) Each of these factors can be split up: → (2x + y)(2x - y)(x + 2y)(x - 2y) –––––––––––––––––––––––
Huge factoring Question?
16x^2-72xy+81y^2 16x² - 36xy - 36xy + 81y² ← Factor by grouping 4x(4x - 9y) - 9y(4x - 8y) (4x - 9y)(4x - 9y) (4x - 9y)² ← Answer x^2(x^2-5x+6)-x^2+5x+6 x²(x² - 5x + 6) - 1(x² - 5x - 6) Do you have this question written correctly? (2y^2+y)-13(2y^2+y)+30 2y(y + 1) - 13*2y(y + 1) + 30 2y(y + 1) - 26y(y + 1) + 30 (y + 1)(2y - 26y) + 30 (y + 1)(-24y) + 30 -24y² - 24y + 30 -6(4y² - 4y + 5) ← Answer (m-2n)^3+(m+2n)^3 ← This is the difference of two cubes, where: (a³ + b³) = (a + b)(a² - ab + b²) [(m - 2n) + (m + 2n)][(m - 2n)² - (m - 2n)(m + 2n) + (m + 2n)²] [2m][(m² - 4mn + 4n²) - (m² - 4n²) + (m² + 4mn + 4n²)] 2m[m² - 4mn + 4n² - m² + 4n² + m² + 4mn + 4n²] 2m[m² + 12n²] ← Answer 27(x-y)^3-(x+y)^3 ← Difference of two cubes, where: a³ - b³ = (a - b)(a² + ab + b²) 3²(x - y)³ - (x + y)³ [3(x - y) - (x + y)]{[3(x - y)]² + 3(x - y)(x + y) + (x + y)²} [3x - 3y - x - y] {[3(x² - 2xy + y²)] + 3(x² - y²) + (x² + 2xy + y²)} [2x - 4y] {3x² - 6xy + 3y² + 3x² - 3y² + x² + 2xy + y²} 2(x - 2y)(7x² - 4xy + y²) ← Answer x^2+2xy+y^2+3x+3y+2 ← Factor by grouping (x² + 2xy + y²) + (3x + 3y) + 2 (x + y)(x + y) + 3(x + y) + 2 (x + y)² + 3(x + y) + 2 ← Let z = (x + y) z² + 3z + 2 z² + 1z + 2z + 2 z(z + 1) + 2(z + 1) (z + 2)(z + 1) ← Replace (x + y) = z (x + y + 2)(x + y + 1) ← Answer
Can someone factorise (x^2-x) y^2+y-(x^2+x)?
Given equation is,(x^2-x) y^2+y-(x^2+x)It is Quadratic equation in y of form a*y^2+b*y+cHere,a= x(x-1) , b= 1 , c= x(x+1); where a!=0Solutions to this equation are,y= [-b+(b^2 - 4ac)]/2a , y= [-b-(b^2 - 4ac)]/2a ;Thus, Substituting values,b^2–4ac= 1- (4x^2)*(x^2–1) -b+ b^2–4ac = (4x^2)(x^2–1)y= (4x^2)(x^2–1)/x(x-1)y= 4x(x+1)Now,Product of roots is c/ac/a= x(x+1)/x(x-1) = (x+1)/(x-1) ;Thus, y*4x(x-1) = (x+1)/(x-1) ;y= (x+1)/ [4x(x-1)]Thus, factors of this equation are,y= 4x(x+1) and y= (x+1)/ [4x(x-1)]List of Formulas:1. a^2 - b^2 = (a+b)(a-b)2. ab-ac = a(b-c)
Find the integrating factor of differential equation ydx + (x^2*y -x)dy = 0?
In Mdx + Ndy = 0 let Δ = ∂M/∂y−∂N/∂x To see if there is an x-only or y-only IF check Δ/N & Δ/M for x-only & y-only M = y, N = x²y−x, Δ = 1 – (2xy−1) = 2(1−xy) → g(x) = Δ/N = −2/x Since this is x-only μ(x) = exp( ∫ g(x)dx ) = 1/x² Read this : http://www.proofwiki.org/wiki/Integratin...
How do you Factor the following expression,[math] x^2-y^2-2x-2y[/math]?
Here is how I would factor the algebra expression.First thing I recognized was the difference between two squares. The more algebra you do, the more you will usually start to recognize patterns like this.I used colors to show the connections between steps.Physics and Math Questions?