Given that K is a constant, if the polynomial X & sup2;, - kxy-3xy-3y & sup2; - 8 does not contain XY term, what is the value of K? How many times and terms is the polynomial?

Given that K is a constant, if the polynomial X & sup2;, - kxy-3xy-3y & sup2; - 8 does not contain XY term, what is the value of K? How many times and terms is the polynomial?

If there is no XY term, then the coefficient of XY is equal to 0. First, we combine the original formula with the similar terms to get X & sup2; - (K + 3) xy-3y & sup2; - 8
So K + 3 = 0. K = - 3
This polynomial is a quadratic trinomial
How to solve (x-1 / 3x-12) - (1 / 4x + 15) = 1 / 2x + 53
The original formula = (2 / 3) x - (1 / 4) x-12-15 = (1 / 2) x + 53
(-1/12)x=80
x=-960
(x-1/3x-12)-(1/4x+15)=1/2x+53
x(1/4-1/3)=80
x=-960
X equals 40 plus minus two root sign 6394
X=-960
Factorization of Polynomials: x ^ 2-4y ^ 2
Thank you very much~
x^2-4y^2=x^2-(2y)^2=（x-2y）（x+2y）
12. X ∈ (1 / 8,1 / 7), find the value of | 1-2x | + | 1-3x | + | 1-4x | +... + | 1-10x |
13.2x ^ 4-x ^ 3-6x ^ 2-x + 2 factorization into (2x-1) Q (x), q (x)
12 because 1 / 80,..., 1-7x > 0,1-8x
Factorization factor X ^ 2-4y ^ 2 + x-2y
The original formula = (x + 2Y) (x-2y) + (x-2y)
=(x-2y)(x+2y+1)
3x ^ 2-4x + 6 = 9 find 2x ^ 2-8 / 3x + 6 =?
3x^2-4x+6=9
3x^2-4x=(9-6)=3
(3x^2-4x)/3=3/3=1
We get x ^ 2-4 / 3x = 1
So 2x ^ 2-8 / 3x = 2
Factorization factor X ^ 2y-4y
x^2y-4y
=y(x^2-4)
=y(x-2)(x+2)
x^2y-4y
=y(x^2-4)
=y(x+2)(x-2)
x²y-4y
=y(x²-4)
=Y (x + 2) (X-2): Thank you very much, too
For any real number x, compare the values of two algebraic expressions 3x3-2x2-4x + 1 and 3x3 + 4x + 10
(3x3-2x2-4x + 1) - (3x3 + 4x + 10) = - 2x2-8x-9 = - 2 (x2 + 4x) - 9 = - 2 [(x + 2) 2-4] - 9 = - 2 (x + 2) 2-1 ＜ 0, that is, (3x3-2x2-4x + 1) - (3x3 + 4x + 10) ＜ 0, (3x3-2x2-4x + 1 ＜ 3x3 + 4x + 10)
Decomposition factor: x ^ 2-x-4y ^ 2 + 2Y
x^2-x-4y^2+2y
=(x^2-4y^2)-(x-2y)
=(x+2y)(x-2y)-(x-2y)
=(x-2y)(x+2y-1)
（x^2-4y^2）－（x－2y)
（x+2y）（x-2y）－（x－2y)
(x-2y)(x+2y-1)
x^2-x-4y^2+2y
=x^2-4y^2-（x-2y）
=（x+2y+(x-2y)-（x-2y）
=(x-2y)(x+2y-1)
x^2-x-4y^2+2y =x^2-4y^2-x+2y =(x^2-4y^2)-(x-2y)=( x-2y)(x+2y)-(x-2y)= (x-2y)( x+2y-1)
2X 10 8 10 3x 10 8 10 4x 10 8 =? Please work out the problem
The answer is 9x + 24