(x2 + 1) = (x2 - i2) with the factors (x+i)(x-i) where i is the square root of (-1).
If you don't allow complex numbers (which is usually the case in school mathematics) then there is no factorisation.
(x + 2)(x + 1)
It is (x+5)(x+1) when factored
It is (x+6)(x+7) when factored
x2+7x x(x+7)
x2 + 4x = x(x + 4)
(5x + 2)(x + 3)
x^2 - 5x - 6 Since the coefficient of 'x^(2)' is '1' in this case. we look at the constant '6' . So write down all the factors of 6' . They are 1,2,3,&6. From these four number we select two numbers that add/subtract to '- So if we say -6 + 1 = -5 5'. The coefficient of 'x' Since 6 - 1 = 5 & 2 + 3 = 5 We need to think a little deeper. So we look at the mathemetical signs , which in this case are both negative (-). Since the sign in front of the '6' is negative, this means that the signs in the brackets are different. NB If it was a possitive both the signs in brackets would be the same either both (+) or both (-). So to have a '-5' we can say -6 + 1 = -5 So setting up the brackets ( x 6 )(x 1) So the signs will be (x - 6)( x + 1) Done!!!! Hope that helps!!!!
x + x = x(1+1)
17
X squared + 7X + 12?
(x - 3)(2x - 7)
Factorise it: (x + 5)(3x - 8)