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 42 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+6x^2+18x-27

Practice makes perfect

We want to write a polynomial function with integral coefficients so that f(x)=0 has the given zeros. -3, 1, -3i 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
-3 x-(-3)
1 x-1
-3i x-(-3i)
3i x-3i
Polynomial f(x)= [x-(-3)] (x-1) [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-(-3)] * (x-1) [x-(-3i)] * (x-3i) After we find these products, we will multiply the obtained expressions.
(x-(-3))(x-1)
(x+3)(x-1)
x(x-1)+3(x-1)
â–Ľ
Distribute x & 3
x^2-x+3(x-1)
x^2-x+3x-3
x^2+2x-3
Let's continue by finding the product of the last two factors.
(x-(-3i))(x-3i)
(x+3i)(x-3i)
x^2-(3i)^2
x^2-3^2 i^2
x^2-9i^2
x^2-9(-1)
x^2+9
Finding the product of these two polynomials will give us the desired polynomial function.
f(x)=(x^2+2x-3)(x^2+9)
f(x)=x^2(x^2+9)+2x(x^2+9)-3(x^2+9)
â–Ľ
Distribute x^2 & 2x & -3
f(x)=x^4+9x^2+2x(x^2+9)-3(x^2+9)
f(x)=x^4+9x^2+2x^3+18x-3(x^2+9)
f(x)=x^4+9x^2+2x^3+18x-3x^2-27
f(x)=x^4+6x^2+2x^3+18x-27
f(x)=x^4+2x^3+6x^2+18x-27