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 15 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^4-3x^3-x^2-27x-90

Practice makes perfect

We want to write a polynomial function with integral coefficients so that f(x)=0 has the given zeros. -2, 5, -3i To do so, recall the Complex Conjugates Theorem for complex zeros.

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
-2 x-(-2)
5 x-5
-3i x-(-3i)
3i x-3i
Polynomial f(x)= [x-(-2)] (x-5) [x-(-3i)](x-3i)

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-(-2)] * (x-5) [x-(-3i)] * (x-3i) After we find these products, we will multiply the obtained expressions.

[x-(-2)](x-5)
(x+2)(x-5)
x(x-5)+2(x-5)
â–¼
Distribute x & 2
x^2-5x+2(x-5)
x^2-5x+2x-10
x^2-3x-10

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

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

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

f(x)=(x^2-3x-10)(x^2+9)
f(x)=x^2(x^2+9)-3x(x^2+9)-10(x^2+9)
â–¼
Distribute x^2 & - 3x & -10
f(x)=x^4+9x^2-3x(x^2+9)-10(x^2+9)
f(x)=x^4+9x^2-3x^3-27x-10(x^2+9)
f(x)=x^4+9x^2-3x^3-27x-10x^2-90
f(x)=x^4-3x^3-x^2-27x-90