a polynomial f(x) whose coefficients are real numbers. Find the remaining zero of f. Give the expression in the standard form a + bi.
Degree 5; zeros: 3, 4, 5, i

Question

a polynomial f(x) whose coefficients are real numbers. Find the remaining zero of f. Give the expression in the standard form a + bi.
Degree 5; zeros: 3, 4, 5, i

zale101 · Answer

factor the zeroes out

f(x)=(x-3),(x-4), (x-5), and for i the factor will be (x - i)(x + i) because every complex numbers which be paired up conjugately

zale101 · Answer

f(x)=(x-3)(x-4)(x-5) (x - i)(x + i) 
factor  (x - i)(x + i)  you'll get (x² + 1) 
f(x)=(x-3)(x-4)(x-5)(x² + 1) 
factor (x-3)(x-4) you'll get x^2-7x+12
(x^2-7x+12)(x-5)(x² + 1) 

now you factor from up here?

anonymous · Answer

x^5 - 12x^4 + 48x^3 -6x^2 + 47x -60 ??

zale101 · Answer

yes :)

anonymous · Answer

how would i express it in a +bi form?

whpalmer4 · Answer

Your "missing" zero is $ -i$, right?  $$a+bi = -i$$Solve for $a,b$

whpalmer4 · Answer

Didn't need to do all that algebra, you're only asked to find the remaining zero and express it as $a+bi$!

anonymous · Answer

sorry i dont really get what the question is asking

whpalmer4 · Answer

Well, I'll explain that in a minute.  First, can you solve that equation I gave you for $a$ and $b$?

anonymous · Answer

no :/

whpalmer4 · Answer

Really?  How about this?

$$a + bx = -x$$Can you solve that for $a$ and $b$?

whpalmer4 · Answer

I hope it is clear that $b = -1$ and $a = 0$...

anonymous · Answer

a=-b

whpalmer4 · Answer

$$0 - 1i = -i$$right?

whpalmer4 · Answer

Just like $$0 - 1x = -x$$

anonymous · Answer

yea

whpalmer4 · Answer

So, $-i$ in $a+bi$ form is just $0-1i$

whpalmer4 · Answer

All the problem is asking you to do is know that in a polynomial with only real coefficients, any complex zeros must come in conjugate pairs, that is to say, $a\pm bi$.  You correctly saw that you had $0+1i$ as a zero, and the problem just wanted you to realize that $0-1i$ must also be a zero.

anonymous · Answer

ohk!!! i get what it asking now! thanks for explaining the question, i havent attempted a question like that before,so i was a bit confused

whpalmer4 · Answer

The answer is kind of hiding in plain sight :-)

mathstudent55 · Answer

A polynomial equation with real coefficients has only real roots and/or complex roots in complex conjugate pairs. Also, a polynomial equation has as many roots as its degree.

You were given a 5th degree polynomial equation with real coefficients. Therefore, there are 5 roots. Since you are given 3 real roots and 1 complex root, you are given 4 roots, so all you are looking for is one more root. Since one of the roots is a complex root, i, it must have a complex conjugate root along with it. The complex conjugate of i is -i. Therefore, the missing root is -i. Now you need to express -i as a complex number in a + bi form which @whpalmer4 has explained above.

anonymous · Answer

thanks alot for the help guys!

mathstudent55 · Answer

You're welcome.

whpalmer4 · Answer

When you have complex roots in conjugate pairs like this, and you multiply the whole thing out, they annihilate each other, leaving only real numbers behind.

For example, say we had a simple polynomial with roots $a+bi, a-bi$:$$P(x) = (x-(a+bi)i)(x-(a-bi)i) = (x-a-bi)(x-a+bi)$$If we expand that, we get$$P(x) = x^2-ax+bix - ax + a^2 -abi -bix +abi - b^2i^2$$$$P(x) = x^2-2ax+a^2-b^2i^2$$But $i^2 =-1$, so we can simplify a bit:
$$P(x) = x^2-2ax+a^2-b^2(-1)$$$$P(x) = x^2-2ax+a^2+b^2$$Now we have no imaginary numbers, only real ones.

whpalmer4 · Answer

Sorry, that first line should be 
$$P(x) = (x-(a+bi))(x-(a-bi)) = (x-a-bi)(x-a+bi)$$Correct after that.

I started with a simpler example, then changed my mind...didn't quite update the first step correctly!