3xsquare+11x-20

Question

3x^{2 +11x-20}

 

Answer 1

Math often teaches us how to go forward and to go backward between equations, expressions, … or other things.

The FOIL (First-Outside-Inside-Last) method goes from:
                (Ax + B)(Cx + D)      to      ACx² + ADx + BCx + BD

Or, in this problem, we want to go backward

          from     ACx² + ADx + BCx + BD     to     (Ax + B)(Cx + D)     

So, the problem starts with    3x² + 11x – 20     which means:
     AC = 3          (the two first numbers multiply giving 3)
     BD = -20      (the last two numbers multiplied give -20, so one must be negative)
     AD + BC  = 11     (which is why you entered ‘middle term breaking’)

      Are there two numbers (one positive, one negative) that multiply giving -20 and
      can produce 11 when multiplied by other numbers and the results added?

Later, you will learn The Binomial Formula which calculates ‘solutions’ to this.

For now,  we can try some:
     Hmm..    (-4)*5 is -20   and  3*5 + (-4)1  = 11 
                     (note: sometimes checking prime factors here helps)

               (3x-4)(x+5)  =  3x² +15x -4x - 20         ( using FOIL)
                            Yep, that’s  3x² +11x -20

Answer 2

If you wish to factor 3x^2+11x-20 then realize the following:

 

A quadratic ax^2+bx+c can not always be factored as (dx+e)(fx+g) where d,e,f and g are real, never mind requiring them to be integers; however, we will assume this quadratic can be factored this way and see if we can deduce the solution.

 

(dx+e)(fx+g)=(df)x^2+(ef+dg)x+eg  so a=df, b=(ef+dg),c=eg

 

In the equation a=3=ed where e and d are integers....the only possible values for f and d are 1 and 3.  We can arbitrarily set f=1 and d=3.

 

(3x+e)(x+g)

 

Now c=-20=eg.  So (e,g) can be (+/-1,-/+20),(+/-20,-/+1),(+/-2,-/+10),(+/-10,-/+2),(+/-4,-/+5),(+/-5,-/+4)

 

b=ef+dg=e+3g=11

 

If we look for the set (e,g) that satisfies e+3g=11 we find (-4,5)...-4+3(5)=-4+15=11

 

So f=1, d=3, e=-4, g=5 and  3x^2+11x-20=(3x-4)(x+5)

 

Check: (3x-4)(x+5)=3x^2+15x-4x-20  OK