Exercise 13 Page 217

Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 3x^2-16x+5 ⇔ 3x^2+(- 16)x+5 We have that a= 3, b=- 16, and c=5. There are now three steps we need to follow in order to rewrite the above expression.

1. Find a c. Since we have that a= 3 and c=5, the value of a c is 3* 5=15.
   
   1. Find factors of a c. Since a c=15, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=- 16, which is negative, those factors will need to be negative so that their sum is negative.

   c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result -3 &-5 &-3 + (-5) & -8 -1 & - 15 & -1 + ( - 15) &- 16

   3. Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 3x^2+(- 16)x+5 ⇔ 3x^2 - 1x - 15x+5

   Finally, we will factor the last expression obtained.

   3x^2-1x-15x+5

   FactorOut

   Factor out x

   (x(3x-1)-15x+5)

   FactorOut

   Factor out - 5

   ( x(3x-1)-5(3x-1) )

   FactorOut

   Factor out (3x-1)

   (x-5)(3x-1)

   Now we can set these factors equal to 0 and solve using the Zero Product Property.

   (x-5)(3x-1)=0

   ZeroProdProp

   Use the Zero Product Property

   lcx-5=0 & (I) 3x-1=0 & (II)

   AddEqn

   (I): LHS+5=RHS+5

   lx=5 3x-1=0

   AddEqn

   (II): LHS+1=RHS+1

   lx=5 3x=1

   DivEqn

   (II): .LHS /3.=.RHS /3.

   lx=5 x= 13

   The solutions to the given equation are x=5 and x= 13.