To factorise the expression (5x^{10}), you can identify the common factors. The expression can be factored as (5 \cdot x^{10}). If you are looking to factor it further, you could express (x^{10}) as ((x^5)^2), but the primary factorization remains (5x^{10} = 5 \cdot x^{10}).
p2-5p-50= (p-10) (p+5)
x(x+5)
To factorise the expression (10x^2 - 15xy), first identify the common factors in both terms. The common factor is (5x). Factoring this out, we get: [ 10x^2 - 15xy = 5x(2x - 3y) ] Thus, the factorised form is (5x(2x - 3y)).
25xy+5yz is 5y(5x+z) when factored
To factorise the expression (10x - 6z), first identify the greatest common factor (GCF) of the coefficients, which is 2. You can then factor out 2 from each term: (10x - 6z = 2(5x - 3z)). Thus, the factorised form is (2(5x - 3z)).
p2-5p-50= (p-10) (p+5)
qwertyuiopasdfjk
(5x + 2)(x + 3)
you factor out the common 5so it would be: 5(x + 2)so that when you multiply back out 5 by x you get 5x and 5 by 2 you get 10
x(x+5)
6x^2 + 5x + 1 = (3x +1) (2x + 1)
1
(4x-3)(x+2)
To factorise the expression (10x^2 - 15xy), first identify the common factors in both terms. The common factor is (5x). Factoring this out, we get: [ 10x^2 - 15xy = 5x(2x - 3y) ] Thus, the factorised form is (5x(2x - 3y)).
25xy+5yz is 5y(5x+z) when factored
Yes and it is 4(5x-7) when factored
5x2 + 9x + 13 has no real factors.