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 16 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^3-4x^2-15x+68

Practice makes perfect

We want to write a polynomial function with integral coefficients so that f(x)=0 has the given zeros. -4, 4+i 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
-4 x-(-4)
4+i x-(4+i)
4-i x-(4-i)
Polynomial f(x)= [x-(-4)] [x-(4+i)] [x-(4-i)]

Let's simplify the polynomial by applying the Distributive Property. For simplicity, we will start by multiplying the first two factors and then we will multiply the result by the last factor.

f(x)=(x-(-4))(x-(4+i))(x-(4-i))
f(x)=(x+4)(x-(4+i))(x-(4-i))
f(x)=(x+4)(x-4-i)(x-4+i)
f(x)=(x(x-4-i)+4(x-4-i))(x-4+i)
â–¼
Distribute x & 4
f(x)=(x^2-4x-ix+4(x-4-i))(x-4+i)
f(x)=(x^2-4x-ix+4x-16-4i)(x-4+i)
f(x)=(x^2-ix-16-4i)(x-4+i)
f(x)=x^2(x-4+i)-ix(x-4+i)-16(x-4+i)-4i(x-4+i)
â–¼
Distribute x^2 & - ix & -16 & -4i
f(x)=x^3-4x^2+ix^2-ix(x-4+i)-16(x-4+i)-4i(x-4+i)
f(x)=x^3-4x^2+ix^2-ix^2+4ix-i^2x-16(x-4+i)-4i(x-4+i)
f(x)=x^3-4x^2+ix^2-ix^2+4ix-i^2x-16x+64-16i-4i(x-4+i)
f(x)=x^3-4x^2+ix^2-ix^2+4ix-i^2x-16x+64-16i-4ix+16i-4i^2
f(x)=x^3-4x^2+ix^2-ix^2+4ix-(-1)x-16x+64-16i-4ix+16i-4(-1)
f(x)=x^3-4x^2+ix^2-ix^2+4ix+1x-16x+64-16i-4ix+16i-4(-1)
f(x)=x^3-4x^2+ix^2-ix^2+4ix+1x-16x+64-16i-4ix+16i+4
f(x)=x^3-4x^2-15x+68