McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
10. Roots and Zeros
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Exercise 41 Page 77

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

f(x)=x^4+2x^3+5x^2+8x+4

Practice makes perfect

We want to write a polynomial function with integral coefficients so that f(x)=0 has the given zeros. -1, -1, 2i To do so, recall the Complex Conjugates Theorem.

Complex Conjugates Theorem

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

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

Root Factor
-1 x-(-1)
-1 x-(-1)
2i x-2i
-2i x-(-2i)
Polynomial f(x)= [x-(-1)] [x-(-1)] (x-2i)[x-(-2i)]

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-(-1)] & * [x-(-1)] (x-2i) & * [x-(-2i)] After we find these products, we will multiply the obtained expressions.

[x-(-1)][x-(-1)]
(x+1)(x+1)
x(x+1)+1(x+1)
â–¼
Distribute x & 1
x^2+x+1(x+1)
x^2+x+x+1
x^2+2x+1

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

(x-2i)[x-(-2i)]
(x-2i)(x+2i)

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

x^2-(2i)^2
x^2-2^2i^2
x^2-4i^2
x^2-4(-1)
x^2+4

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

f(x)=(x^2+2x+1)(x^2+4)
f(x)=x^2(x^2+4)+2x(x^2+4)+1(x^2+4)
â–¼
Distribute x^2 & 2x & 1
f(x)=x^4+4x^2+2x(x^2+4)+1(x^2+4)
f(x)=x^4+4x^2+2x^3+8x+1(x^2+4)
f(x)=x^4+4x^2+2x^3+8x+x^2+4
f(x)=x^4+5x^2+2x^3+8x+4
f(x)=x^4+2x^3+5x^2+8x+4