How do you factor completely #4c^2 - 12c + 9#?

Question

Genesis Allen · Accepted Answer

# (2c - 3 ) (2c -3) # is the factorised form of the expression.

#4c^2 - 12c + 9 # We can Split the Middle Term of this expression to factorise it. In this technique, if we have to factorise an expression like #xc^2 + yc + z#, we need to think of 2 numbers such that: #N_1*N_2 = x*z = 4 * 9 = 36# AND #N_1 +N_2 = y = -12# After trying out a few numbers we get #N_1 = -6# and #N_2 =-6# #-6*-6 = 36#, and #(-6)+(-6)= -12# #4c^2 - 12c + 9 = 4c^2 - 6c - 6c + 9 # # = 2c(2c -3) -3 (2c - 3)# #(2c - 3 )# is a common factor to each of the terms # = (2c - 3 ) (2c -3) #

Eli Assouline · Answer

undefined To factor completely $ 4c^2 - 12c + 9 $, you can use the quadratic formula. The quadratic formula states that for a quadratic equation $ ax^2 + bx + c = 0 $, the solutions for $ x $ are given by $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $. In this case, $ a = 4 $, $ b = -12 $, and $ c = 9 $. Plugging these values into the quadratic formula, you get $ x = \frac{{-(-12) \pm \sqrt{{(-12)^2 - 4(4)(9)}}}}{{2(4)}} $. Simplifying inside the square root, you get $ x = \frac{{12 \pm \sqrt{{144 - 144}}}}{{8}} $. Since the discriminant is zero, the solutions are equal. Thus, $ x = \frac{{12}}{{8}} = \frac{{3}}{{2}} $. Therefore, the factored form of the quadratic equation is $ (2c - 3)^2 $.