Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Chapter Review
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Exercise 55 Page 350

If a is a root of P(x)=0, then (x-a) is a factor of P(x). If P(x) is a polynomial with real coefficients, then the complex roots of P(x)=0 occur in conjugate pairs.

P(x)=x^4-4x^3-10x^2+68x-80

Practice makes perfect

We want to write a polynomial function with rational coefficients so that P(x)=0 has the given roots. 3+i, 2, and - 4 To do so, recall the Conjugate Root Theorem for complex roots.

Conjugate Root Theorem

If P(x) is a polynomial with real coefficients, then the complex roots of P(x)=0 occur in conjugate pairs.

This theorem states that if a + bi is a complex root, then a - bi is also a root. Additionally, recall that if a is a root of P(x)=0, then (x-a) is a factor of P(x).

Root Factor
3+i x-(3+i)
3-i x-(3-i)
2 x-2
-4 x-(-4)
Polynomial P(x)= (x-(3+i)) (x-(3-i)) (x-2) (x-(-4))

Let's simplify the polynomial by applying the Distributive Property. For simplicity, we will start by multiplying the first two factors and the last two factors separately. (x-(3+i)) * (x-(3-i)) (x-2) * (x-(-4)) After we find these products, we will multiply the obtained expressions.

(x-(3+i))(x-(3-i))
(x-3-i)(x-3+i)
x(x-3+i)-3(x-3+i)-i(x-3+i)
â–¼
Distribute x & - 3 & i
x^2-3x+ix-3(x-3+i)-i(x-3+i)
x^2-3x+ix-3x+9-3i-i(x-3+i)
x^2-3x+ix-3x+9-3i-ix+3i-i^2
x^2-3x+ix-3x+9-3i-ix+3i-(-1)
x^2-3x+ix-3x+9-3i-ix+3i+1
x^2-6x+10

Let's continue by finding the product of the last two factors.

(x-2)(x-(-4))
(x-2)(x+4)
x(x+4)-2(x+4)
â–¼
Distribute x & - 2
x^2+4x-2(x+4)
x^2+4x-2x-8
x^2+2x-8

Finding the product of these two polynomials will give us the desired polynomial function.

P(x)=(x^2-6x+10)(x^2+2x-8)
P(x)=x^2(x^2+2x-8)-6x(x^2+2x-8)+10(x^2+2x-8)
â–¼
Distribute x^2 & - 6x & 10
P(x)=x^4+2x^3-8x^2-6x(x^2+2x-8)+10(x^2+2x-8)
P(x)=x^4+2x^3-8x^2-6x^3-12x^2+48x+10(x^2+2x-8)
P(x)=x^4+2x^3-8x^2-6x^3-12x^2+48x+10x^2+20x-80
P(x)=x^4-4x^3-10x^2+68x-80