Find all the zeros of f(x)=x3−6x2+25x−68 given that 1−4i is a zero.

Answer 1

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

Answer 2

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

= x² - 2x + 1 +16. (Remember i^2 = -1)

= x² - 2x +17

Now we divide (long division) the polynomial, f(x)=x³−6x²+ 25x−68, by x² - 2x +17

x³−6x²+ 25x−68 ⁄ x² - 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