Checkby substituting, in turn, 0, and 1 into the original equation. The solution
set is
(b)
Factor
Zero-factor propertyor or Solve each equation.
Checkeach solution to verify that the solution set is E NOW TRY
13 , 4, 5F.
x= x= 5 x= 4
1
3
3 x- 1 = 0or x- 5 = 0or x- 4 = 0
1 3 x- 121 x- 521 x- 42 = 0 x^2 - 9 x+20.
13 x- 121 x^2 - 9 x+ 202 = 0
5 - 1, 0, 1 6.
- 1,
z= 0 or z=- 1 or z= 1
6 z= 0 or z + 1 = 0 or z - 1 = 0
SECTION 6.5 Solving Quadratic Equations by Factoring 397
CAUTION In Example 6(b),it would be unproductive to begin by multiplying
the two factors together. The zero-factor property requires the productof two or more
factors to equal 0. Always consider first whether an equation is given in an appro-
priate form for the zero-factor property to apply.
Solving an Equation Requiring Multiplication before FactoringSolve
The zero-factor property requires the productof two or more factors to equal 0.
Multiply.
Combine like terms.
Standard form
Factor.
Zero-factor propertySolve each equation.Checkthat the solution set is E- 32 , 2F. NOW TRY
x=-
3
2
or x= 2
2 x+ 3 =0or x- 2 = 0
12 x+ 321 x- 22 = 0
2 x^2 - x- 6 = 0
3 x^2 +x=x^2 + 2 x+ 6
3 x^2 +x=x^2 + 2 x+ 1 + 5
13 x+ 12 x= 1 x+ 122 + 5
13 x+ 12 x= 1 x+ 122 + 5.
EXAMPLE 7
1 x+ 122 = 1 x+ 121 x+ 12Complete solution available
on the Video Resources on DVD
6.5 EXERCISES
Concept Check In Exercises 1–5, fill in the blank with the correct response.
1.A quadratic equation in xis an equation that can be put into the form
2.The form is called form.
3.If a quadratic equation is in standard form, to solve the equation we should begin by
attempting to the polynomial.
4.The equation is not a quadratic equation, because.
5.If a quadratic equation has then mustbe a solution
because is a factor of the polynomial.ax^2 +bx+c= 0 c=0,x^3 +x^2 +x= 0ax^2 +bx+c= 0=0.
NOW TRY
EXERCISE 6
Solve each equation.
(a)
(b)
12 a^2 - 5 a- 122 = 0
13 a- 12 #
3 x^3 - 27 x= 0NOW TRY
EXERCISE 7
Solve.
x 14 x- 92 = 1 x- 222 + 24NOW TRY ANSWERS
- (a) (b)
- E-^73 , 4F
5 - 3, 0, 3 (^6) E- 23 ,^13 , 4F