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 53 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^3+3x^2+25x+75

Practice makes perfect

We want to write a polynomial function with rational coefficients so that P(x)=0 has the given roots. - 3 and 5i 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
5i x-5i
-5i x-(- 5i)
- 3 x-(-3)
Polynomial P(x)= (x-5i) (x-(-5i)) (x-(-3))

Let's simplify the polynomial by applying the Distributive Property. For simplicity, we will start by multiplying the first two factors together, and then multiply this product by the last factor.

(x-5i)(x-(-5i))
(x-5i)(x+5i)
x(x+5i)-5i(x+5i)
â–¼
Distribute x & -5i
x^2+5ix-5i(x+5i)
x^2+5ix-5ix-25i^2
x^2+5ix-5ix-25(-1)
x^2+5ix-5ix+25
x^2+25

Let's continue by multiplying this product by the third factor.

(x^2+25)(x+3)
x^2(x+3)+25(x+3)
â–¼
Distribute x^2 & 25
x^3+3x^2+25(x+3)
x^3+3x^2+25x+75