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 44 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^3-9x^2+77x+150

Practice makes perfect

We want to write a polynomial function with integral coefficients so that f(x)=0 has the given zeros. -2, -3, 4-3i 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
-2 x-(-2)
-3 x-(-3)
4-3i x-(4-3i)
4+3i x-(4+3i)
Polynomial f(x)= [x-(-2)] [x-(-3)] [x-(4-3i)][x-(4+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-(-3)] [x-(4-3i)] & * [x-(4+3i)] After we find these products, we will multiply the obtained expressions.

[x-(-2)][x-(-3)]
(x+2)(x+3)
x(x+3)+2(x+3)
â–¼
Distribute x & 2
x^2+3x+2(x+3)
x^2+3x+2x+6
x^2+5x+6

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

[x-(4-3i)][x-(4+3i)]
(x-4+3i)(x-4-3i)
((x-4)+3i)((x-4)-3i)
(x-4)^2-(3i)^2
x^2-8x+16-(3i)^2
x^2-8x+16-3^2i^2
x^2-8x+16-9i^2
x^2-8x+16-9(-1)
x^2-8x+16+9
x^2-8x+25

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

f(x)=(x^2+5x+6)(x^2-8x+25)
f(x)=x^2(x^2-8x+25)+5x(x^2-8x+25)+6(x^2-8x+25)
â–¼
Distribute x^2 & 5x & 6
f(x)=x^4-8x^3+25x^2+5x(x^2-8x+25)+6(x^2-8x+25)
f(x)=x^4-8x^3+25x^2+5x^3-40x^2+125x+6(x^2-8x+25)
f(x)=x^4-8x^3+25x^2+5x^3-40x^2+125x+6x^2-48x+150
f(x)=x^4-3x^3-9x^2+77x+150