No
(2x-1)(2x-1) = 4x^2 -4x + 1
x^2 + 4x + 5 cannot be factored because its discriminant is less than zero.
Remember both 16 & 25 are squared numbers. 16 = 4^2 & 25 = 5^2 Hence we can write (4x)^2 - (5y)^2 Remember two squared terms with a NEGATIVE Between them will factor. ( 4x - 5y)(4x + 5y) Note the difference signs. NNB Two squared terms with a positive (+) between them DOES NOT factor.
4x^(2) - 8x '4' is common to both '4' & '8' . So tak outr '4'. Hence 4(x^(2) - 2x) Inside the brackets 'x' is common to both 'x^(2) & '2x'. So take out 'x'. Hence 4x(x - 2) Done , fully factored.
Given 4x2+4x-1 Using the formula for the roots of quadratic equation, (-b +/-./b2-4ac)/2a the roots for the above will be (-4+/-4./2)/8 = (-1+/-./2)/2 Hence the two roots are (-1+./2)/2 and (-1-./2)/2 As the roots are irrational, there is no possibility of getting factors.
4(x^2 + x + 3)
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!!!!!
4(x+y)^2
(2x-1)(2x-1) = 4x^2 -4x + 1
2(x - 1)(2x + y)
2(x^2 + 2)(x + 3)
2(4x - 3)(x + 3)
x^2 + 4x + 5 cannot be factored because its discriminant is less than zero.
The answer is (2x^2+3)(4x+1)
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
4x2+4x-8 4(x2+x-2) and (4x+8)(x-1)
(2x - 9)(2x - 9) or (2x - 9)2