Question 59424
1>  m^2 + 4m - 21 = (m+7) (m-3)





2> 2x^3 +5x^2 - 12x = 0
Removing x as the common factor,  the above equation becomes,
x(2x^2 + 5x - 12) = 0
==> x [ 2x^2 +8x - 3x - 12 ] = 0 [splitting the middle term]
==> x[ 2x(x+4) -3(x+4) ] = 0 [taking out the common factor]
==> x(x+4) (2x-3) = 0
==> either x = 0   or x+4 = 0  or 2x - 3 = 0

==> x = 0  or x = -4  or x = 3/2         [2x-3 = 0==>2x = 3 or x = 3/2]








3>  2y^2 = -3y + 35

Adding 3y-35 to both the sides,

   2y^2 + 3y - 35 = 0
==> 2y^2 + 10y - 7y - 35 = 0  [splitting the middle term]
==> 2y(y + 5) - 7(y + 5) = 0
==>(y + 5) (2y - 7) = 0
==> y+5 = 0   or 2y - 7 = 0
==> y = -5  or  y = 7/2




4>  (4x-3)(4x+3)
   = (4x)^2 - 3^2  [as (a+b)(a-b) = a^2 - b^2]
   = 16x^2 - 9


Good Luck!!!