4x^4 or 4x*4x*4x*4x A formula would be (x^a)(x^b)= x^a+b x^a) / (x^b)= x^a-b
x2 - 4x - 5 = zero( x - 5 ) ( x + 1 ) = zerox - 5 = zero ...... X = 5x + 1 = Zero ...... X = -1
16x^2 - 25 Recognise that both 16 and 25 are squared numbers, 16 = 4^(2) & 25 = 5^(2) Substituting 4^(2)x^(2) - 5^(2) = > (4x)^(2) - 5^(2) We now have two squared terms with a negative between them. NB two squared terms with positive does NOT factor. Factoring (4x - 5)(4x + 5)
x = ¼x^4 + 4x + k where k can be any number
x = -½
It is: (x-5)(x+1) when factored
no, because xx=x squared, and x squared is not linear
4x^4 or 4x*4x*4x*4x A formula would be (x^a)(x^b)= x^a+b x^a) / (x^b)= x^a-b
(4x + 5)(x - 3)
4x2 - x - 5 = 0 4x2 + 4x - 5x - 5 = 0 4x(x + 1) - 5(x + 1) = 0 (4x - 5)(x + 1) = 0 4x - 5 = 0 or x + 1 = 0 4x = 5 or x = -1 x = 5/4 o x = -1
x^2 + 4x + 5 cannot be factored because its discriminant is less than zero.
Find factors of -5 that can be added to make -4, in this case -5 and 1, so: (x - 5)(x + 1) = x2 - 4x - 5
x2 - 4x - 5 = zero( x - 5 ) ( x + 1 ) = zerox - 5 = zero ...... X = 5x + 1 = Zero ...... X = -1
4x squared
x2 + 4x = x(x+4)
x2 + 4x = x(x + 4)
4x^2 = 2 * 2 * x * x