Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
3. Rational Functions and Their Graphs
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Exercise 53 Page 523

Factor by applying the difference of cubes.

x=2/3

Practice makes perfect
To solve the given equation by factoring, we will apply the difference of two cubes. a^3-b^3 ⇔ (a-b)(a^2+ab+b^2) Let's start by factoring out the greatest common factor.
54x^3-16=0
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Factor left-hand side
2( 27x^3-8 ) =0
2( 3^3x^3-2^3 ) =0
2( (3x)^3-2^3 ) =0

a^3-b^3 = (a-b)(a^2+ab+b^2)

2(3x-2)( (3x)^2+3x* 2+2^2 ) =0
2(3x-2)( (3x)^2+6x+2^2 ) =0
2(3x-2)( 3^2x^2+6x+2^2 ) =0
2(3x-2)( 9x^2+6x+4 ) =0
We have rewritten the left-hand side as a product of three factors. Now, we will apply the Zero Product Property to solve the equation. Note that since 2≠ 0, we will not consider it when applying the property.
(3x-2)( 9x^2+6x+4 ) =0
lc3x-2=0 & (I) 9x^2+6x+4=0 & (II)
l3x=2 9x^2+6x+4=0
lx= 23 9x^2+6x+4=0
From Equation (I), we found that one solution is x= 23. To find other solutions, we will solve Equation (II). Note that since this is a quadratic equation, we will use the Quadratic Formula. ax^2+bx+c=0 ⇔ x=- b±sqrt(b^2-4ac)/2a To do so, we first need to identify a, b, and c. 9x^2+ 6x+ 4=0 We can see above that a= 9, b= 6, and c= 4. Let's substitute these values into the formula and solve for x.
x=- b±sqrt(b^2-4ac)/2a
x=- 6±sqrt(6^2-4( 9)( 4))/2( 9)
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Solve for x
x=- 6±sqrt(36-4(9)(4))/2(9)
x=- 6±sqrt(36-36(4))/2(9)
x=- 6±sqrt(36-144)/2(9)
x=- 6±sqrt(- 108)/2(9)
x=- 6±sqrt(- 108)/18
x=- 6± sqrt(36* (-1)* 3)/18
x=- 6± sqrt(36)* sqrt(- 1)* sqrt(3)/18
x=- 6± 6sqrt(- 1) sqrt(3)/18
x=- 6± 6i sqrt(3)/18
x=6( - 1± i sqrt(3) )/18
x=- 1± i sqrt(3)/3
We found that the solutions for the quadratic equation are x= - 1± isqrt(3)3. Notice that they are not real solutions, thus the only real solution is 23. Real solution x=2/3