Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 7.2
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Exercise 164 Page 362

Practice makes perfect
a

To use the Zero Product Property to solve this equation, we set each factor on the left-hand side equal to 0 and solve for x.

x(2x+1)(3x-5)=0
lcx=0 & (I) 2x+1=0 & (II) 3x-5=0 & (III)
lx=0 2x=- 1 3x-5=0
lx=0 x=- 12 3x-5=0
lx=0 x=- 12 3x=5
lx_1=0 x_2=- 12 x_3= 53

b

Here we cannot use the Zero Product Property, because one of the equation's sides does not equal 0. Therefore, we must first rewrite the equation so that one side equals zero.

(x-3)(x-2)=12
x^2-2x-3x+6=12
x^2-5x+6=12
x^2-5x-6=0

To factor this equation, we have to identify two numbers, a and b, whose product equals the equation's constant - 6, and whose sum equals the x-term's coefficient, - 5. These numbers we substitute into an equation in the following format.

(x+a)(x+b)=0 Let's search for our numbers by identifying factors whose product equals -6, and then check if the factor's sum equals -5. c|c|c|c|c Product & ab & a+b & Sum & - 6 & -2(3) & - 2+3 & 1 & * - 6 & -3(2) & - 3+2 & - 1 & * - 6 & -1(6) & - 1+6 & 5 & * - 6 & -6(1) & - 6+1 & - 5 & ✓ When a=1 and b=- 6, we will have factored the equation. With this information, we can solve the equation with the Zero Product Property.

(x+a)(x+b)=0
(x+ 1)(x+( - 6))=0
(x+1)(x-6)=0
x+1=0 & (I) x-6=0 & (II)
x=- 1 x-6=0
x_1=- 1 x_2=6

The solutions are x_1=- 1 and x_2=6.

c

As long as we write the equation so that one side consist of only factors and the other side is 0, we can solve it by the Zero Product Property. Below we see an example.

(3x-3)^(factor 1) (x-20)^(factor 2) (2x-10)^(factor 3)=0 Let's solve this equation with the Zero Product Property.

(3x-3)(x-20)(2x-10)=0
lc3x-3=0 & (I) x-20=0 & (II) 2x-10=0 & (III)
l3x=3 x-20=0 2x-10=0
lx=1 x-20=0 2x-10=0
lx=1 x=20 2x-10=0
lx=1 x=20 2x=10
lx=1 x=20 x=5