(3x + 1)(x + 5)
(x + 5)(3x + 1)
Re=Write as 16x^(3) + 12x^(2) + 4x Take out the common factor of '4' Hence 4(4x^(3) + 3x^)2) + x) Take out the common factor of 'x' Hence 4x( 4x^(2) + 3x + 1) The answer!!!!!
NOTE: xx can be written as x^2 i.e. x squared 3x^2 + 16x + 5 = 3 x^2 + x + 15x + 5 (Since 16x = x + 15x) = x ( 3x + 1 ) + 5 ( 3x + 1 ) = ( x + 5 ) ( 3x + 1 ) Therefore the factors are ( x + 5 ) and ( 3x + 1 )
(5x - 3)(3x + 5)
(3x - y+ 2)(3x + y + 2)
(x + 5)(3x + 1)
Re=Write as 16x^(3) + 12x^(2) + 4x Take out the common factor of '4' Hence 4(4x^(3) + 3x^)2) + x) Take out the common factor of 'x' Hence 4x( 4x^(2) + 3x + 1) The answer!!!!!
NOTE: xx can be written as x^2 i.e. x squared 3x^2 + 16x + 5 = 3 x^2 + x + 15x + 5 (Since 16x = x + 15x) = x ( 3x + 1 ) + 5 ( 3x + 1 ) = ( x + 5 ) ( 3x + 1 ) Therefore the factors are ( x + 5 ) and ( 3x + 1 )
(5x - 3)(3x + 5)
It is: 3x*(x+7)
(3x - y+ 2)(3x + y + 2)
3(x2 + 3x + 1)
48x2+139+63 = (3x+7)(16x+9) when factored and worked out with the help of the quadratic equation formula
(3x + 3y)(x + y) = 3(x + y)2
If that's -18y2, the answer is (3x - 2y)(x + 9y)
(6x + t)(3x + t)
3x(2x + 1)