Can #y= x^2-3x-28 # be factored? If so what are the factors ?

1 Answer
Dec 7, 2015

Answer:

#x^2 - 3x - 28 = (x-7)(x+4)#

Explanation:

Luckily, your #x^2# term already has the factor #1# in front - this makes your computation easier.

I would like to suggest the following method:

If your factorization exists, it would look like this:

#x^2 - 3x - 28 = (x + a)(x + b)#

#color(white)(xxxxxxxxx) = x^2 + (a+b)x + a*b#

So, your goal is to find #a# and #b# with the following properties:

  • #a + b = -3#
  • #color(white)(i)a xx b color(white)(i) = -28#

It is always easier to start with the multiplication. Since #-28# is negative, one of the two numbers needs to be positive and the other one negative.
Possible integer pairs of #a# and #b# are:

  • #(-4) xx 7 color(white)(xx)# or #color(white)(xx) 4 xx (-7)#
  • #(-2) xx 14 color(white)(xii)# or #color(white)(xx)2 xx (-14)#
  • #(-1) xx 28 color(white)(xii)# or #color(white)(xx)1 xx (-28)#

It is easy to check that the correct pair is #4# and #-7# since #4 + (- 7) = -3# and #4 * (-7) = -28#.

Thus, your factorization is

#x^2 - 3x - 28 = (x-7)(x+4)#.

#color(white)(x)#

By the way, this also means that the solutions for the equation

#x^2 - 3x - 28 = 0#

are #-a# and #-b#, so #x = 7# or #x = -4#.