Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
3. Section 8.3
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Exercise 124 Page 412

Use an area model to divide the polynomial.

x_1=5
x_2=2+sqrt(5)
x_3=2-sqrt(5)

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Since the graph has a factor of (x-5), we can write the following equation. (x-5) (other factor)=x^3-9x^2+19x+5 Let's use an area model to find the other factor. The first term of the other factor must be x^2. It must be this in order for the product of the factor's first terms to equal x^3. ( x-5) ( x^2......)= x^3-9x^2+19x+5

With this information, we can begin creating our area model's first column.

In the original expression we have the term -9x^2. Since one tile of our area model contains - 5x^2, we must add - 4x^2 to get a sum of -9x^2. With this information, we can identify the second term of our factor and the contents of the area model's second column.

Again, examining the original expression, we see the term 19x. Since one tile of our area model contains 20x, we must add - x to get a sum of 19x. With this information, we can identify the third term of our factor and thereby, the contents of the third column.

If we add all of the terms contained within the area model, the sum should equal the expression. x^3+(- 5x^2)+(-4x^2)+20x+(- x)+5 ⇓ x^3-9x^2+19x+5 Now we know that the other factor is (x^2-4x-1). With this information, we can rewrite the original expression. (x-5) (x^2-4x-1) We can find the remaining roots by equating the second factor with 0 and solving for x by using the Quadratic Formula.
x^2-4x-1=0
x=-( - 4) ± sqrt(( - 4)^2-4 * 1( - 1))/2 * 1
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Solve for x
x=4 ± sqrt((- 4)^2-4 * 1(- 1))/2 * 1
x=4 ± sqrt(16+4)/2
x=4 ± sqrt(20)/2
x=4 ± sqrt(4* 5)/2
x=4 ± sqrt(4)sqrt(5)/2
x=4 ± 2sqrt(5)/2
lcx=.(4+2sqrt(5)) /2. & (I) x=.(4-2sqrt(5)) /2. & (II)
lx_1=2+sqrt(5) x_2=2-sqrt(5)
The two remaining roots are x=2+sqrt(5) and x=2-sqrt(5), which means their corresponding factors are (x-(2+sqrt(5))) and (x-(2-sqrt(5))). Now we can also fully factor our expression. (x-5)(x-(2+sqrt(5)))(x-(2-sqrt(5)))