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 43 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-x^3-20x^2+50x

Practice makes perfect

We want to write a polynomial function with integral coefficients so that f(x)=0 has the given zeros. 0, -5, 3+i 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
0 x-0
-5 x-(-5)
3+i x-(3+i)
3-i x-(3-i)
Polynomial f(x)= (x-0) [x-(-5)] [x-(3+i)][x-(3-i)]

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-0) & * [x-(-5)] [x-(3+i)] & * [x-(3-i)] After we find these products, we will multiply the obtained expressions.

(x-0)[x-(-5)]
(x-0)(x+5)
x(x+5)
x^2+5x

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

[x-(3+i)][x-(3-i)]
(x-3-i)(x-3+i)
((x-3)-i)((x-3)+i)

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

(x-3)^2-i^2
x^2-6x+9-i^2
x^2-6x+9-(-1)
x^2-6x+9+1
x^2-6x+10

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

f(x)=(x^2+5x)(x^2-6x+10)
f(x)=x^2(x^2+5x)-6x(x^2+5x)+10(x^2+5x)
â–¼
Distribute x^2 & - 6x & 10
f(x)=x^4+5x^3-6x(x^2+5x)+10(x^2+5x)
f(x)=x^4+5x^3-6x^3-30x^2+10(x^2+5x)
f(x)=x^4+5x^3-6x^3-30x^2+10x^2+50x
f(x)=x^4-x^3-20x^2+50x