How do you factor #2x^{2}-13x+24#?

2 Answers
Aug 11, 2017

#f(x) = 2x^2 - 13x + 24#
#D = b^2 - 4ac = 169 - 192 = - 23#
Since D < 0 and D is not a perfect square, this f(x) can't be factored.

Aug 11, 2017

See below

Explanation:

In this case, the trinomial #a!=1# so we can use bottom-up method

Step 1: Multiply "a" to "c" (#a=2 , b=24#)
#x^2-13x+(24xx2#)
#x^2-13x+48#

Step 2: Now we can factor by using "magic X" which finds 2 numbers that can multiply equal 48 and add up to #-13#
#(x-16)(x+3)#

Step 3: Divide the numbers by the leading coefficient (#2#)
#(x-16/2)(x+3/2)#

Step 4: If the numbers divide “evenly” then this is the factor. If the numbers canNOT divide evenly, then pull the divisor up to be the leading coefficient of the x term
#(x-8)(2x+3)#
If you want to find x value, you can use Zero Product Property (set each factor equal to zero)

#x-8=0#, #x=8#
#2x+3=0#, #2x=-3#, #x=-3/2#