other zeroes are the conjugate of 1-4i which is 1+4i, and the real root 4

the cubic function can be factored as (x-4)(x^2-2x +17)

set the factors = 0 and solve for x: x=4, x = 1+4i, 1-4i

Shayleigh S.

asked • 01/05/21
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other zeroes are the conjugate of 1-4i which is 1+4i, and the real root 4

the cubic function can be factored as (x-4)(x^2-2x +17)

set the factors = 0 and solve for x: x=4, x = 1+4i, 1-4i

Ritu L. answered • 01/05/21

Tutor

New to Wyzant
20+ yr exp, result orientated Math Expert up to Calculus

Hello Shayleigh,

Zeros means all x-intercepts. And you are given 1 zero/x-intercept, which is a complex/imaginary number.

Complex number zero will always have other zero of its conjugate. So automatically you have the 2nd zero, 1+4i.

So 2 zeros are 1-4i and 1+4i or 1±4i.

Since its a cubic polynomial, we need to find the 3rd zero.

We can divide the polynomial by the **product** of 2 imaginary zeros in factor form.

Product of {x-(1-4i)} and {x-(1+4i)} is {x-(1-4i)} {x-(1+4i)}

After rearranging terms,

= {(x-1) + 4i} {(x-1) -4i}

= (x-1)^{2} - (4i)^{2}

= x^{2} - 2x + 1 +16. (Remember i^2 = -1)

= x^{2} - 2x +17

Now we divide (__long division__) the polynomial, **f(x)=x**^{3}**−6x**^{2}**+ 25x−68, **by x^{2} - 2x +17

**x**^{3}**−6x**^{2}**+ 25x−68 ⁄ **x^{2} - 2x +17

**= x - 4**

So we got the 3rd zero of 4.

So 3 zeros are **1 ± 4i, and 4.**

Please do let me know if you need more help and we can schedule a lesson here.

Thanks

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