Solve each equation by factoring

Question

1) (2b-3) (b+2)=0

 

2) r^2+2r-48=0

 

3) x^2+7x=8

 

4) 8a^2 - 24a=0

 

 

Answer 1

# 1 is already factored.
      So either 2b-3 = 0 or b+2 = 0 and these are the 2 answers for x.

#2 needs to be factored.
     You need to look for 2 numbers that

     multiply to give -48 and add to give +2

  The easy but long way to do this is to list all the factors of -48

     and find two that add to 2.

     Sometimes it is simpler to just think about it and you will find the numbers.

 

  For this they are 8 and -6

     so to factor it gives (x+8)(x-6)

  So in this either x+8 = 0  or  x-6 = 0 gives the 2 answers for x.

 

#3 You need to change it by subtracting 8 from both sides.

      x^2 + 7x + -8 = 0

     so you need to factor it by finding the two numbers

     that multiply to -8 and add to 7

     find those numbers and put them into

      (x+number1) (x+number2) = 0

      then solve x+number1=0  and  x+number2=0

     and these will be the 2 answers for x

 

#4 needs to be factored by taking a common factor out.

     8A^2 - 24A = 0

     becomes   8(A^2 + 3A) = 0

     one of these factors must be 0 for the equation to be = 0

     but 8 cannot be = 0  so  (A^2 + 3A) must be = 0

     so lets factor A^2 + 3A = 0 by factoring out an A

     we get  A (A+3)

     there for either A = 0   or   A+3 = 0

     those are the 2 answers for x