# Use the given zero to find the remaining zeros of the function. f(x)=x^3-4x^2+36x-144; zero: 6 i

Asked on by katie3848

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

If `6i` is a zero of the function, then `(x-6i)` will divide `f(x)`. Even better, `f(x)` has only real coefficients. (No complex ones, such as `2+3i` ). This means that all the complex zeros come in "complex conjugate pairs." So, if `5+4i` is a zero, then `5-4i` is also a zero. In our case, that means `-6i` is a zero of the function, so `(x+6i)` will divide our function. So:

`(x-6i)(x+6i)=x^2-6ix+6ix-36i^2=x^2+36` divides `f(x)`

From here we can use polynomial division:

(Hopefully this comes out ok:)

____________________
x^2 + 0x + 36 | x^3 - 4x^2 + 36x -144

What can you multiply x^2 by to get x^3? A: 1x, so just as in regular division:

1x
________________________
x^2 + 0x + 36 | x^3 - 4x^2 + 36x -144

Now multiply:

1x
________________________
x^2 + 0x + 36 | x^3 - 4x^2 + 36x -144
x^3 + 0x^2 + 36x
--------------------

Subtract:

1x
________________________
x^2 + 0x + 36 | x^3 - 4x^2 + 36x -144
x^3 + 0x^2 + 36x
------------------
-4x^2 + 0x -144

What can you multiply x^2 by to get -4x^2? A: -4

1x -4
________________________
x^2 + 0x + 36 | x^3 - 4x^2 + 36x -144
x^3 + 0x^2 + 36x
------------------
-4x^2 + 0x -144

Multiply:

1x -4
________________________
x^2 + 0x + 36 | x^3 - 4x^2 + 36x -144
x^3 + 0x^2 + 36x
-----------------
-4x^2 + 0x -144
-4x^2 + 0x -144
----------------

Subtract, and you get 0, which means there is no remainder. We knew this would be the case, because x^2 + 36 must evenly divide f(x)

Thus we have:

`f(x)=(x-6i)(x+6i)(x-4)=(x^2+36)(x-4)`

And the zeros are:

6i, -6i, 4

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