factorise (a-m)(b+n)
Answers
Answer:
Factorization of Polynomials
Factorization is the decomposition of an expression into a product of its factors.
The following are common factorizations.
For any positive integer nn,
a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).
a
n
−b
n
=(a−b)(a
n−1
+a
n−2
b+…+ab
n−2
+b
n−1
).
In particular, for n=2n=2, we have a^2-b^2=(a-b)(a+b)a
2
−b
2
=(a−b)(a+b).
For an odd positive integer nn,
a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).
a
n
+b
n
=(a+b)(a
n−1
−a
n−2
b+…−ab
n−2
+b
n−1
).
a^2 \pm 2ab + b^2 = (a\pm b)^2a
2
±2ab+b
2
=(a±b)
2
x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
(ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)(ax+by)
2
+(ay−bx)
2
=(a
2
+b
2
)(x
2
+y
2
), (ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2)(ax−by)
2
−(ay−bx)
2
=(a
2
−b
2
)(x
2
−y
2
)
x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x)x
2
y+y
2
z+z
2
x+x
2
z+y
2
x+z
2
y+2xyz=(x+y)(y+z)(z+x)
Factorization often transforms an expression into a form that is more easily manipulated algebraically, has easily recognizable solutions, and gives rise to clearly defined relationships.
Step-by-step explanation:
Find all ordered pairs of integer solutions (x,y)(x,y) such that 2^x+ 1 = y^22
x
+1=y
2
.
We have 2^x = y^2-1 = (y-1)(y+1)2
x
=y
2
−1=(y−1)(y+1). Since the factors (y-1)(y−1) and (y+1)(y+1) on the right hand side are integers whose product is a power of 2, both (y-1)(y−1) and (y+1)(y+1) must be powers of 2. Furthermore, their difference is
(y+1)-(y-1)=2,
(y+1)−(y−1)=2,
implying the factors must be y+1 = 4y+1=4 and y-1 = 2y−1=2. This gives y=3y=3, and thus x=3x=3. Therefore, (3, 3)(3,3) is the only solution.
Answer:
This value cannot be factorised....
Anyway hi friend.......