Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
3. Section 8.3
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Exercise 154 Page 423

Practice makes perfect
a Examining the graph, we notice that it intercepts the x-axis at x=2.

The Remainder Theorem tells us that if p(c)=0, then (x-c) is a factor. Therefore, since p(2)=0, it must be that (x-2) is a factor.

b Since (x-2) is a factor, we can write the following equation.
(x-2) (other factor)=x^3+4x^2+x-26 Let's use an area model to find the other factor. The first term of the other factor must be x^2. It has to be this in order for the product of the factor's first terms to equal x^3. ( x-2) ( x^2......)= x^3+4x^2+x-26

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

In the original expression, we have the term 4x^2. Since one tile of our area model contains - 2x^2, we must add 6x^2 to get a sum of 4x^2. With this information we can identify the second term of our factor, and thereby the contents of the area model's second column.

Again, examining the original expression, we find the term x. Since one tile of our area model contains - 12x, we must add 13x to get a sum of x. 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 right-hand side of the polynomial. x^3+(- 2x^2)+6x^2+(- 12x)+13x-26 ⇓ x^3+4x^2+x-26 Now we know that another factor is (x^2+6x+13). With this information, we can rewrite the original expression. (x-2) (x^2+6x+13)=x^3+4x^2+x-26 To find the remaining roots, we should set the second degree factor equal to 0 and solve for x with the Quadratic Formula.
x^2+6x+13=0
x=- 6 ± sqrt(6^2-4( 1)( 13))/2( 1)
â–Ľ
Solve for x
x=-6 ± sqrt(36-52)/2
x=-6 ± sqrt(- 16)/2
x=-6 ± isqrt(16)/2
x=-6 ± i 4/2
x=-6 ± 4i/2
lcx=.(- 6+4i) /2. & (I) x=.(- 6-4i) /2. & (II)
lx_1=- 3+2i x_2=- 3-2i
The two complex roots are x=- 3+2i and x=- 3-2i. Now we can identify all of the factors of p(x). p(x)=(x-2)(x-(- 3+2i))(x-(- 3-2i))